The Central Idea
A trigonometric equation is an equation containing one or more trigonometric functions of an unknown angle.
For example,
\[ \sin x=\frac12 \]asks for every angle \(x\) whose sine equals \(1/2\).
Because trigonometric functions are periodic, such equations often have infinitely many solutions.
Identity versus Equation
A trigonometric identity is true throughout its common domain.
A trigonometric equation is true only for selected inputs.
For example,
\[ \sin^2x+\cos^2x=1 \]is an identity, while
\[ \sin x=\frac12 \]is an equation to solve.
What Counts as a Solution?
A solution is an angle that makes the original equation true.
A candidate is not accepted merely because it appears during algebraic manipulation.
The final check is always substitution into the original equation.
Solutions on an Interval
Some problems restrict the answer to an interval such as
\[ 0\leq x<2\pi. \]In that case, list only the solutions inside the specified interval.
Periodic repetitions outside the interval are not included.
General Solutions
If no finite interval is specified, the goal is often to describe every real solution.
Periodicity makes this possible using integer parameters such as
\[ k\in\mathbb Z. \]For example,
\[ \sin x=0 \]has the general solution
\[ \boxed{x=k\pi,\qquad k\in\mathbb Z}. \]Start by Isolating the Trigonometric Function
Many equations become familiar after ordinary algebra isolates the trigonometric expression.
For example,
\[ 2\sin x-1=0 \]becomes
\[ \sin x=\frac12. \]The remaining work is then trigonometric rather than algebraic.
Solving \(\sin x=a\)
If
\[ -1\leq a\leq1, \]first identify a reference or principal angle.
Then use the signs of sine to locate all relevant quadrants.
Sine is positive in Quadrants I and II and negative in Quadrants III and IV.
Example: Solve \(\sin x=1/2\) on One Revolution
Solve
\[ \sin x=\frac12 \]for
\[ 0\leq x<2\pi. \]The reference angle is
\[ \frac{\pi}{6}. \]Sine is positive in Quadrants I and II.
Therefore,
\[ \boxed{ x=\frac{\pi}{6}, \quad \frac{5\pi}{6} }. \]General Sine Solutions
If
\[ \sin x=\sin\alpha, \]then the two basic solution families are
\[ \boxed{x=\alpha+2\pi k} \]and
\[ \boxed{x=\pi-\alpha+2\pi k}, \]where
\[ k\in\mathbb Z. \]General Solution for \(\sin x=1/2\)
From the one-revolution solutions,
\[ \frac{\pi}{6} \qquad\text{and}\qquad \frac{5\pi}{6}, \]the complete solution is
\[ \boxed{ x=\frac{\pi}{6}+2\pi k } \]or
\[ \boxed{ x=\frac{5\pi}{6}+2\pi k, \qquad k\in\mathbb Z }. \]Solving \(\cos x=a\)
Cosine is positive in Quadrants I and IV and negative in Quadrants II and III.
Once a reference angle is known, use the sign of the requested cosine value to determine the appropriate quadrants.
Example: Solve \(\cos x=-\sqrt2/2\)
Solve on
\[ 0\leq x<2\pi. \]The reference angle is
\[ \frac{\pi}{4}. \]Cosine is negative in Quadrants II and III.
Therefore,
\[ \boxed{ x=\frac{3\pi}{4}, \quad \frac{5\pi}{4} }. \]General Cosine Solutions
If
\[ \cos x=\cos\alpha, \]then
\[ \boxed{x=2\pi k\pm\alpha}, \]where
\[ k\in\mathbb Z. \]This compact form expresses the reflection symmetry of cosine.
Solving \(\tan x=a\)
Tangent has period
\[ \pi, \]so once one solution \(\alpha\) is known, all solutions are
\[ \boxed{x=\alpha+k\pi,\qquad k\in\mathbb Z}. \]Tangent is positive in Quadrants I and III and negative in Quadrants II and IV.
Example: Solve \(\tan x=1\)
The principal/reference angle is
\[ \frac{\pi}{4}. \]Tangent repeats every \(\pi\).
Therefore,
\[ \boxed{ x=\frac{\pi}{4}+k\pi, \qquad k\in\mathbb Z }. \]Example: Tangent on One Revolution
Solve
\[ \tan x=-\sqrt3 \]for
\[ 0\leq x<2\pi. \]The reference angle is
\[ \frac{\pi}{3}. \]Tangent is negative in Quadrants II and IV.
Therefore,
\[ \boxed{ x=\frac{2\pi}{3}, \quad \frac{5\pi}{3} }. \]Inverse Trigonometric Functions Give Principal Values
When an exact unit-circle value is unavailable, use an inverse trigonometric function to obtain a principal angle.
For example,
\[ \sin x=0.37 \]gives
\[ \alpha=\arcsin(0.37). \]This is only one principal value. The complete solution must still account for sine's symmetry and period.
Example: Non-Special Sine Value
Solve
\[ \sin x=0.37 \]for
\[ 0\leq x<2\pi. \]The principal angle is approximately
\[ \alpha=\arcsin(0.37)\approx0.379. \]Since sine is positive in Quadrants I and II, the second solution is
\[ \pi-\alpha\approx2.763. \]Thus,
\[ \boxed{x\approx0.379,\ 2.763}. \]Equations with Reciprocal Functions
Reciprocal equations can often be rewritten in terms of sine, cosine, or tangent.
For example,
\[ \sec x=2 \]is equivalent to
\[ \cos x=\frac12. \]Then solve the familiar cosine equation.
Example: Solve \(\csc x=-2\)
Rewrite:
\[ \sin x=-\frac12. \]On
\[ 0\leq x<2\pi, \]sine is negative in Quadrants III and IV, with reference angle \(\pi/6\).
Therefore,
\[ \boxed{ x=\frac{7\pi}{6}, \quad \frac{11\pi}{6} }. \]Example: Solve \(\cot x=-1\)
Since
\[ \cot x=\frac1{\tan x}, \]the equation is equivalent to
\[ \tan x=-1. \]Thus,
\[ \boxed{ x=-\frac{\pi}{4}+k\pi, \qquad k\in\mathbb Z }. \]Use Algebra before Trigonometry
Consider
\[ 3\cos x+1=0. \]First isolate cosine:
\[ 3\cos x=-1, \] \[ \cos x=-\frac13. \]Only then use inverse cosine and quadrant reasoning.
Quadratic-Looking Trigonometric Equations
Some equations are quadratic in a trigonometric function.
For example,
\[ 2\sin^2x-\sin x-1=0 \]has the same algebraic structure as
\[ 2u^2-u-1=0 \]if
\[ u=\sin x. \]Example: Factor a Trigonometric Quadratic
Solve
\[ 2\sin^2x-\sin x-1=0. \]Factor:
\[ (2\sin x+1)(\sin x-1)=0. \]Therefore,
\[ \sin x=-\frac12 \]or
\[ \sin x=1. \]Complete the Factored Example
On
\[ 0\leq x<2\pi, \]the equation
\[ \sin x=-\frac12 \]gives
\[ x=\frac{7\pi}{6}, \quad \frac{11\pi}{6}. \]The equation
\[ \sin x=1 \]gives
\[ x=\frac{\pi}{2}. \]Thus,
\[ \boxed{ x=\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} }. \]Zero-Product Reasoning
If an equation has been written as
\[ F(x)G(x)=0, \]then at least one factor must be zero:
\[ F(x)=0 \]or
\[ G(x)=0. \]This familiar algebraic principle is extremely useful in trigonometric equations.
Example: Factor a Common Trigonometric Factor
Solve
\[ \sin x\cos x-\sin x=0. \]Factor:
\[ \sin x(\cos x-1)=0. \]Therefore,
\[ \sin x=0 \]or
\[ \cos x=1. \]Do Not Divide Away Possible Solutions
In
\[ \sin x\cos x-\sin x=0, \]dividing both sides by \(\sin x\) would produce
\[ \cos x-1=0. \]But this division assumes
\[ \sin x\neq0. \]Any solutions with \(\sin x=0\) would be lost.
Factoring is safer.
Using Trigonometric Identities
Identities can convert an equation into a more solvable form.
The most common substitutions include
\[ \sin^2x+\cos^2x=1, \] \[ 1+\tan^2x=\sec^2x, \]and
\[ 1+\cot^2x=\csc^2x. \]Example: Use the Pythagorean Identity
Solve
\[ \sin^2x=1-\cos x. \]Replace
\[ \sin^2x \]with
\[ 1-\cos^2x. \]Then
\[ 1-\cos^2x=1-\cos x. \]Continue the Identity Example
Subtract \(1\):
\[ -\cos^2x=-\cos x. \]Multiply by \(-1\):
\[ \cos^2x=\cos x. \]Move all terms to one side:
\[ \cos^2x-\cos x=0. \]Factor:
\[ \cos x(\cos x-1)=0. \]Finish the Identity Example
Thus,
\[ \cos x=0 \]or
\[ \cos x=1. \]On
\[ 0\leq x<2\pi, \]the solutions are
\[ \boxed{ x=0, \frac{\pi}{2}, \frac{3\pi}{2} }. \]Equations Involving Both Sine and Cosine
If an equation contains both sine and cosine, possible strategies include:
- factoring;
- using a Pythagorean identity;
- dividing only after separately checking zero cases;
- rewriting tangent as \(\sin x/\cos x\);
- rewriting everything in one trigonometric function.
Example: Convert to Tangent
Suppose
\[ \sin x=\cos x \]and \(\cos x\neq0\).
Divide by \(\cos x\):
\[ \frac{\sin x}{\cos x}=1. \]Therefore,
\[ \tan x=1. \]Hence,
\[ \boxed{ x=\frac{\pi}{4}+k\pi }. \]Why the Division Was Safe in That Example
If
\[ \cos x=0, \]the original equation
\[ \sin x=\cos x \]would require
\[ \sin x=0 \]at the same angle, which never occurs.
Therefore no valid solutions are lost by dividing by \(\cos x\) in this particular equation.
Equations with a Transformed Angle
The trigonometric input may itself contain an algebraic expression:
\[ \sin(2x)=\frac12. \]A useful strategy is to temporarily treat the entire argument as one angle.
Let
\[ u=2x. \]Then solve
\[ \sin u=\frac12 \]before converting back to \(x\).
Example: Solve \(\sin(2x)=1/2\)
Let
\[ u=2x. \]Then
\[ \sin u=\frac12. \]Thus,
\[ u=\frac{\pi}{6}+2\pi k \]or
\[ u=\frac{5\pi}{6}+2\pi k. \]Convert Back to \(x\)
Since \(u=2x\),
\[ 2x=\frac{\pi}{6}+2\pi k \]gives
\[ x=\frac{\pi}{12}+\pi k. \]The second family gives
\[ x=\frac{5\pi}{12}+\pi k. \]Therefore,
\[ \boxed{ x=\frac{\pi}{12}+\pi k \quad\text{or}\quad x=\frac{5\pi}{12}+\pi k }. \]Interval Restrictions with Transformed Arguments
Suppose
\[ \sin(2x)=\frac12 \]must be solved for
\[ 0\leq x<2\pi. \]Then
\[ 0\leq2x<4\pi. \]The doubled argument covers two full sine cycles, so more solutions appear than in the basic equation \(\sin x=1/2\).
Solutions of the Transformed Example on \(0\leq x<2\pi\)
From
\[ x=\frac{\pi}{12}+\pi k \]and
\[ x=\frac{5\pi}{12}+\pi k, \]choose integer values giving \(0\leq x<2\pi\):
\[ \boxed{ x= \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12} }. \]Equations with a Phase Shift
Consider
\[ \cos\left(x-\frac{\pi}{3}\right)=1. \]Let
\[ u=x-\frac{\pi}{3}. \]Since
\[ \cos u=1 \]when
\[ u=2\pi k, \]we obtain
\[ \boxed{ x=\frac{\pi}{3}+2\pi k }. \]Equations with a Negative Argument
Even–odd identities can simplify signs.
Since
\[ \sin(-x)=-\sin x, \]the equation
\[ \sin(-x)=\frac12 \]becomes
\[ \sin x=-\frac12. \]Equations That Have No Real Solutions
Sine and cosine outputs must lie between \(-1\) and \(1\).
Therefore,
\[ \sin x=1.4 \]and
\[ \cos x=-2 \]have no real solutions.
Always check the possible range before doing unnecessary work.
Boundary Values
Equations such as
\[ \sin x=1 \]have only one solution per sine period:
\[ \boxed{ x=\frac{\pi}{2}+2\pi k }. \]Similarly,
\[ \sin x=-1 \]gives
\[ \boxed{ x=\frac{3\pi}{2}+2\pi k }. \]Special Zero Equations
Several common zero equations have compact general forms:
\[ \boxed{\sin x=0\iff x=k\pi}, \] \[ \boxed{ \cos x=0 \iff x=\frac{\pi}{2}+k\pi }, \]and
\[ \boxed{\tan x=0\iff x=k\pi}. \]Domain Restrictions in Equations
An equation may contain expressions that are undefined at certain angles.
For example,
\[ \tan x=\frac{\sin x}{\cos x} \]requires
\[ \cos x\neq0. \]A candidate where the original expression is undefined can never be a solution.
Extraneous Solutions
An extraneous solution is a candidate produced by a transformation that does not satisfy the original equation.
Trigonometric equations can acquire extraneous candidates through:
- squaring both sides;
- multiplying by expressions that may be zero;
- changing domains through simplification;
- using algebraically nonreversible steps.
Example: Squaring Can Create Candidates
Suppose
\[ \sin x=\cos x. \]Squaring gives
\[ \sin^2x=\cos^2x. \]But the squared equation is also true when
\[ \sin x=-\cos x. \]Squaring has therefore weakened the original sign condition.
Any solutions found after squaring must be checked in the original equation.
Graphical Interpretation of a Trigonometric Equation
The equation
\[ \sin x=\frac12 \]can be interpreted as asking where the graphs
\[ y=\sin x \]and
\[ y=\frac12 \]intersect.
This explains visually why multiple solutions occur.
Graphing as a Check
A graph can help:
- estimate the number of solutions on an interval;
- check approximate locations;
- identify missed periodic repetitions;
- detect impossible answers.
Exact algebraic and unit-circle reasoning should still be used when exact solutions are required.
Numerical Solutions
Not every trigonometric equation has elementary exact solutions.
For example,
\[ \sin x=0.43 \]normally leads to decimal approximations unless an exact inverse form is acceptable.
In such cases, state the requested precision clearly.
Inverse Form Can Be Exact
Even when no familiar special angle exists, the expression
\[ \arcsin(0.43) \]is an exact symbolic description of the principal angle.
A decimal is an approximation to that exact value.
Periodicity Determines the Final Pattern
The period of the function determines how the solution families repeat.
Sine and cosine use shifts by
\[ 2\pi. \]Tangent and cotangent use shifts by
\[ \pi. \]When the argument contains \(Bx\), the corresponding period in \(x\) is compressed or stretched.
General Solution with a Linear Argument
Consider
\[ \tan(3x)=1. \]Since
\[ \tan u=1 \]gives
\[ u=\frac{\pi}{4}+k\pi, \]we have
\[ 3x=\frac{\pi}{4}+k\pi. \]Therefore,
\[ \boxed{ x=\frac{\pi}{12}+\frac{k\pi}{3} }. \]Factoring a Difference of Squares
Consider
\[ \sin^2x-\cos^2x=0. \]Factor:
\[ (\sin x-\cos x)(\sin x+\cos x)=0. \]Thus,
\[ \sin x=\cos x \]or
\[ \sin x=-\cos x. \]Complete the Difference-of-Squares Example
The first equation gives
\[ \tan x=1, \]while the second gives
\[ \tan x=-1. \]Together, the solutions may be expressed as
\[ \boxed{ x=\frac{\pi}{4}+\frac{k\pi}{2}, \qquad k\in\mathbb Z }. \]Solving with a Substitution
For
\[ 4\cos^2x-4\cos x+1=0, \]let
\[ u=\cos x. \]Then
\[ 4u^2-4u+1=0. \]Factor:
\[ (2u-1)^2=0. \]Therefore,
\[ \cos x=\frac12. \]Complete the Substitution Example
On
\[ 0\leq x<2\pi, \]cosine equals \(1/2\) in Quadrants I and IV:
\[ \boxed{ x=\frac{\pi}{3}, \quad \frac{5\pi}{3} }. \]When a Quadratic Root Is Impossible
Suppose algebra produces
\[ \sin x=2. \]Since
\[ -1\leq\sin x\leq1, \]that algebraic branch contributes no real trigonometric solutions.
Range restrictions can eliminate roots immediately.
Exact Solutions Should Stay Exact
If a solution is
\[ \frac{5\pi}{6}, \]do not replace it unnecessarily with a decimal.
Exact radian forms preserve the unit-circle structure and make periodic generalizations easier to see.
Degree-Measure Equations
The same methods work in degrees.
For example,
\[ \sin\theta=\frac12 \]has solutions
\[ 30^\circ \]and
\[ 150^\circ \]during one revolution.
The general solutions repeat by
\[ 360^\circ. \]General Degree Solutions
For sine,
\[ \sin\theta=\sin\alpha \]gives
\[ \boxed{ \theta=\alpha+360^\circ k } \]or
\[ \boxed{ \theta=180^\circ-\alpha+360^\circ k }. \]Tangent repeats every
\[ 180^\circ. \]Calculator Mode Matters
If the equation is being solved in radians, inverse-trigonometric calculations should be performed in radian mode.
If the expected answers are in degrees, use degree mode or convert explicitly.
A mode mismatch can produce a numerically plausible but contextually wrong answer.
Verify after Approximation
If a solution is rounded, substitute it into the original equation and expect approximate rather than exact equality.
The residual difference should be consistent with the stated rounding.
Common Error: Returning Only the Inverse-Function Output
If
\[ \sin x=\frac12, \]writing only
\[ x=\arcsin\left(\frac12\right) = \frac{\pi}{6} \]misses the Quadrant-II solution and all periodic repetitions.
Common Error: Using \(2\pi\) for Tangent's Repetition
Tangent repeats after
\[ \boxed{\pi}, \]not merely after \(2\pi\).
Therefore,
\[ \tan x=a \]usually has a general solution of the form
\[ x=\alpha+k\pi. \]Common Error: Forgetting the Given Interval
A general solution may be mathematically correct but fail the requested format if the problem asks only for
\[ 0\leq x<2\pi. \]Always read the domain or interval before presenting the final answer.
Common Error: Dividing by a Trigonometric Expression Too Soon
Dividing by
\[ \sin x \]can remove solutions for which
\[ \sin x=0. \]Before dividing by a variable expression, either factor first or separately analyze the zero case.
Common Error: Ignoring Function Range
Equations such as
\[ \cos x=1.2 \]have no real solutions.
Recognizing the range
\[ [-1,1] \]can settle the equation immediately.
Common Error: Forgetting to Divide the Entire Angle Equation
If
\[ 2x=\frac{\pi}{6}+2\pi k, \]dividing by \(2\) gives
\[ x=\frac{\pi}{12}+\pi k. \]The periodic term must be divided as well.
Common Error: Assuming Squaring Is Reversible
Squaring removes sign information.
Therefore candidates produced after squaring must be checked in the original unsquared equation.
A Reliable Procedure for Basic Trigonometric Equations
- Identify the requested interval or whether a general solution is required.
- Use ordinary algebra to isolate the trigonometric function.
- Check whether the requested value lies in the function's range.
- Find a reference or principal angle.
- Use quadrant signs to find all solutions in one period.
- Apply periodicity to construct the general solution if needed.
- Restrict the solutions to the requested interval.
- Verify in the original equation.
A Reliable Procedure for Factorable Equations
- Move all terms to one side.
- Factor algebraically.
- Apply the zero-product property.
- Solve each resulting trigonometric equation separately.
- Combine the solution sets.
- Remove duplicates.
- Check all solutions against the original equation and interval.
A Reliable Procedure for Equations Using Identities
- Identify which trigonometric functions appear.
- Choose an identity that rewrites the equation in fewer function types.
- Simplify using ordinary algebra.
- Factor or substitute if the result is quadratic-like.
- Solve the resulting basic trigonometric equations.
- Preserve all domain restrictions introduced by the original expression.
- Verify every candidate.
A Reliable Procedure for Transformed Arguments
- Treat the complete argument, such as \(2x-\pi/3\), as one temporary angle.
- Solve the basic trigonometric equation for that angle.
- Write the complete periodic families before solving back for \(x\).
- Undo the algebraic transformation.
- Apply any interval restriction to \(x\).
- Verify representative or all interval solutions.
Practice 1: Basic Sine Equation
Solve
\[ \sin x=\frac{\sqrt3}{2} \]for
\[ 0\leq x<2\pi. \]Practice 2: Basic Cosine Equation
Solve
\[ \cos x=-\frac12 \]for
\[ 0\leq x<2\pi. \]Practice 3: General Tangent Equation
Solve
\[ \tan x=-1 \]for all real \(x\).
Practice 4: Algebra before Trigonometry
Solve
\[ 2\cos x+1=0 \]on
\[ 0\leq x<2\pi. \]Practice 5: Reciprocal Function
Solve
\[ \sec x=-2 \]on
\[ 0\leq x<2\pi. \]Practice 6: Factorable Equation
Solve
\[ \sin x(\cos x+1)=0 \]on
\[ 0\leq x<2\pi. \]Practice 7: Trigonometric Quadratic
Solve
\[ 2\cos^2x-\cos x-1=0 \]on
\[ 0\leq x<2\pi. \]Practice 8: Use an Identity
Solve
\[ \sin^2x=\cos^2x \]for all real \(x\).
Practice 9: Transformed Argument
Solve
\[ \cos(2x)=0 \]on
\[ 0\leq x<2\pi. \]Practice 10: Phase Shift
Solve
\[ \sin\left(x-\frac{\pi}{4}\right)=0 \]for all real \(x\).
Practice 11: No Real Solution
Solve
\[ 3\sin x=4 \]over the real numbers.
Practice 12: Identify the Error
A student solves
\[ \sin x(\cos x-1)=0 \]by dividing both sides by \(\sin x\) and concludes that
\[ \cos x=1 \]gives every solution.
Explain the error.
Practice Answers
Practice 1
The reference angle is
\[ \frac{\pi}{3}. \]Sine is positive in Quadrants I and II:
\[ \boxed{ x=\frac{\pi}{3}, \quad \frac{2\pi}{3} }. \]Practice 2
The reference angle is
\[ \frac{\pi}{3}. \]Cosine is negative in Quadrants II and III:
\[ \boxed{ x=\frac{2\pi}{3}, \quad \frac{4\pi}{3} }. \]Practice 3
A principal solution is
\[ -\frac{\pi}{4}. \]Tangent repeats every \(\pi\):
\[ \boxed{ x=-\frac{\pi}{4}+k\pi, \qquad k\in\mathbb Z }. \]Practice 4
\[ 2\cos x+1=0 \]gives
\[ \cos x=-\frac12. \]Therefore,
\[ \boxed{ x=\frac{2\pi}{3}, \quad \frac{4\pi}{3} }. \]Practice 5
\[ \sec x=-2 \]means
\[ \cos x=-\frac12. \]Thus,
\[ \boxed{ x=\frac{2\pi}{3}, \quad \frac{4\pi}{3} }. \]Practice 6
Use the zero-product property:
\[ \sin x=0 \]or
\[ \cos x=-1. \]On \(0\leq x<2\pi\),
\[ \sin x=0 \]gives
\[ x=0,\pi. \]The second equation gives \(x=\pi\), already included.
Therefore,
\[ \boxed{x=0,\pi}. \]Practice 7
Factor:
\[ (2\cos x+1)(\cos x-1)=0. \]Thus,
\[ \cos x=-\frac12 \]or
\[ \cos x=1. \]Therefore,
\[ \boxed{ x=0, \frac{2\pi}{3}, \frac{4\pi}{3} }. \]Practice 8
Rewrite as
\[ \sin^2x-\cos^2x=0. \]Factor:
\[ (\sin x-\cos x)(\sin x+\cos x)=0. \]This gives tangent values \(1\) and \(-1\), so
\[ \boxed{ x=\frac{\pi}{4}+\frac{k\pi}{2}, \qquad k\in\mathbb Z }. \]Practice 9
Cosine is zero when
\[ 2x=\frac{\pi}{2}+k\pi. \]Therefore,
\[ x=\frac{\pi}{4}+\frac{k\pi}{2}. \]Restricting to \(0\leq x<2\pi\) gives
\[ \boxed{ x= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} }. \]Practice 10
Sine is zero when its argument is \(k\pi\):
\[ x-\frac{\pi}{4}=k\pi. \]Therefore,
\[ \boxed{ x=\frac{\pi}{4}+k\pi, \qquad k\in\mathbb Z }. \]Practice 11
\[ 3\sin x=4 \]requires
\[ \sin x=\frac43. \]But sine cannot exceed \(1\).
Therefore,
\[ \boxed{\text{no real solution}}. \]Practice 12
Dividing by \(\sin x\) assumes
\[ \sin x\neq0. \]But any angle satisfying
\[ \sin x=0 \]already makes the original product equal to zero.
Those solutions are lost by the division.
The correct approach is to keep the factorization:
\[ \sin x=0 \]or
\[ \cos x=1. \]What to Remember
- A trigonometric equation is true only for selected angles; an identity is true throughout its common domain.
- First isolate or simplify the trigonometric expression using ordinary algebra.
- Use unit-circle values and reference angles for exact solutions.
- Inverse trigonometric functions return principal values, not automatically every solution.
- Sine and cosine repeat every \(2\pi\); tangent and cotangent repeat every \(\pi\).
- Quadrant signs determine the additional solutions associated with a reference angle.
- Factoring and the zero-product property are often safer than dividing by a trigonometric expression.
- Identities can rewrite equations into a single trigonometric function.
- For transformed inputs such as \(2x-\pi/3\), solve for the entire angle first and then solve back for \(x\).
- Check the range of sine and cosine before attempting impossible equations.
- Squaring and other nonreversible steps may introduce extraneous candidates.
- Always respect the requested interval and verify candidates in the original equation.
Where We Go Next
Trigonometric equations combine algebra, identities, inverse functions, unit-circle geometry, and periodicity into one problem-solving framework.
These same ideas recur whenever periodic models must be inverted, crossing times must be found, phase conditions must be solved, or trigonometric relationships are used in later calculus and applied mathematics.