AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsTrigonometric Equations, Inequalities & IdentitiesEquations & Inequalities with Sinx, Cosx and Tanx

Equations & Inequalities with Sinx, Cosx and Tanx (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Solving simple trigonometric equations

What are simple trigonometric equations?

A simple trigonometric equation is one of the form
- $\sin θ = k, \cos θ = k, or \tan θ = k$
  - where $k$ is a constant
The goal is to find all values of $θ$ in a specified solution interval that make the equation true
Because trigonometric functions are periodic
- there are usually infinitely many solutions to a trigonometric equation across all real numbers
- but a question typically restricts the solution interval
  - e.g. $0 \leq θ < 2 π$

How can I find an initial solution?

The first step is to find an initial solution
- i.e. any one specific value of $θ$ that satisfies the equation
For values of $k$ corresponding to special angles (multiples of $\frac{π}{6}$ or $\frac{π}{4}$ )
- you can use your knowledge of exact trigonometric values from the unit circle
  - E.g. for $\sin θ = \frac{\sqrt{3}}{2}$
  - recognize that $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$
    - so $θ = \frac{π}{3}$ is an initial solution
On a calculator part of the exam
- you can use an inverse trigonometric function on your graphing calculator
  - E.g. for $\sin θ = 0.4$
    - an initial solution is $θ = \sin^{- 1} (0.4) \approx 0.4115$

How can I find other solutions from the initial solution?

Once one solution is known, additional solutions in the interval can be found using the symmetry and periodicity of the trigonometric functions
For sine equations ( $\sin θ = k$ )
- If $θ_{0}$ is one solution, then $π - θ_{0}$ is another solution
  - by symmetry of the sine graph about $θ = \frac{π}{2}$ in the unit circle (see diagram below)
- All other solutions can then be obtained
  - by adding or subtracting integer multiples of $2 π$ (the period) to the first two solutions
- Note that subtracting $2 π$ from $π - θ_{0}$ gives $- π - θ_{0}$
  - If $θ_{0}$ is in the interval $[- π, 0]$ , this automatically gives another solution in $[- π, 0]$
- Note that adding $2 π$ to $π - θ_{0}$ gives $3 π - θ_{0}$
  - If $θ_{0}$ is in the interval $[π, 2 π]$ , this automatically gives another solution in $[π, 2 π]$
For cosine equations ( $\cos θ = k$ )
- If $θ_{0}$ is one solution, then $- θ_{0}$ is another solution
  - by symmetry of the cosine graph about $θ = 0$ in the unit circle (see diagram below)
  - This also corresponds with cosine being an even function
    - i.e. $\cos (- θ) = \cos θ$ in general
- All other solutions can then be obtained
  - by adding or subtracting integer multiples of $2 π$ (the period) to the first two solutions
- Note that adding $2 π$ to $- θ_{0}$ gives $2 π - θ_{0}$
  - If $θ_{0}$ is in the interval $[0, 2 π]$ , this automatically gives another solution in $[0, 2 π]$
For tangent equations ( $\tan θ = k$ ):
- All other solutions differ from the principal solution by integer multiples of $π$
  - since tangent has period $π$

What is the general process for solving a trigonometric equation in a given interval?

Start by rearranging the equation (if needed) so that it has the form $\sin θ = k$ , $\cos θ = k$ , or $\tan θ = k$
- If the equation involves a trigonometric function appearing more than once (e.g. $2 \cos^{2} θ = \cos θ$ ), use algebra (typically factoring) to break it into simpler equations
  - $2 \cos^{2} θ - \cos θ = 0 \Rightarrow \cos θ (2 \cos θ - 1) = 0 \Rightarrow \cos θ = 0 or \cos θ = \frac{1}{2}$
Next find an initial solution
- using exact values or a calculator
Then use symmetry and periodicity
- to find all other solutions in the specified interval
Finally check that all solutions are within the given interval
- and discard any that are not

Examiner Tips and Tricks

When a question gives a specific solution interval (e.g. $0 \leq θ < 2 π$ or $- π \leq θ \leq π$ ), be careful to include all solutions within that interval, and only those.

A common error is to find the principal solution and stop, missing other solutions that come from symmetry or periodicity

Sketching the unit circle (or the relevant trig graph) and marking the solutions visually is a reliable way to check that you have found all of them.

How can the symmetry properties be visualized?

The unit circle is the most reliable tool for visualizing how many solutions exist and where they are
For $\sin θ = k$ (with $- 1 < k < 1$ , $k \neq 0$ )
- there are two solutions in any interval of length $2 π$
- located symmetrically across the vertical line $θ = \frac{π}{2}$
  - If $\sin θ = k$ , then $\sin (π - θ) = k$ is also true

Unit circle diagram showing angles θ and π-θ. It includes points P(cos θ, sin θ) and (cos(π-θ), sin(π-θ)), with the identity sin θ = sin(π-θ). — **Symmetry of the sine function in the unit circle**

For $\cos θ = k$ (with $- 1 < k < 1$ , $k \neq 0$ )
- there are two solutions in any interval of length $2 π$
- located symmetrically across the horizontal axis (i.e. $θ = 0$ )
  - If $\cos θ = k$ , then $\cos (- θ) = k$ and $\cos (2 π - θ) = k$ are also true

Unit circle with angle θ, showing cosine symmetry. Point P at (cos θ, sin θ). Equation: cos θ = cos(−θ) = cos(2π − θ). — **Symmetry of the cosine function in the unit circle**

For $\tan θ = k$
- there is one solution in any interval of length $π$
- and all other solutions are integer multiple of $π$ apart
  - If $\tan θ = k$ , then it's also true that $\tan (θ - π) = k$ , $\tan (θ + π) = k$ , etc.

How are 'hidden quadratic' trigonometric equations solved?

Some trigonometric equations are quadratics in disguise
- They have the form of a quadratic equation
  - but with a trigonometric function in place of the variable
- For example $2 \cos^{2} x + 7 \cos x + 3 = 0$
  - has the same structure as the algebraic equation $2 y^{2} + 7 y + 3 = 0$

These equations can be solved using a substitution
E.g. to solve $2 \cos^{2} x + 7 \cos x + 3 = 0$ in the interval $[0, 2 π)$
- Let $y$ (or another variable) stand for the trigonometric function
  - I.e. $y = \cos x$
  - The equation becomes a standard quadratic in $y$
    - $2 y^{2} + 7 y + 3 = 0$
- Solve the quadratic in $y$ by factoring or using the quadratic formula
  - $(2 y + 1) (y + 3) = 0$ , giving $y = - \frac{1}{2}$ or $y = - 3$
- Substitute the trigonometric function back in
  - $\cos x = - \frac{1}{2}$ or $\cos x = - 3$
- Some solutions may need to be rejected
  - For sine and cosine, only values with $- 1 \leq y \leq 1$ are valid
    - any other value of $y$ corresponds to no solution and must be discarded
    - So in this case $\cos x = - 3$ must be rejected
  - For tangent, all real values of $y$ are valid
- $\cos x = - \frac{1}{2}$ has initial solution $x = \frac{2 π}{3}$
  - which means that, by symmetry, $x = 2 π - \frac{2 π}{3} = \frac{4 π}{3}$ is also a solution
- So the only solutions to $2 \cos^{2} x + 7 \cos x + 3 = 0$ in $[0, 2 π)$ are $x = \frac{2 π}{3}, \frac{4 π}{3}$

Worked Example

Find all values of $x$ , for $0 \leq x < 2 π$ , that satisfy the equation $2 \sin^{2} x = \sin x$ .

Answer

Rearrange the equation so one side is zero

$2 \sin^{2} x - \sin x = 0$

Factor out the common factor of $\sin x$

$\sin x (2 \sin x - 1) = 0$

This gives two simpler equations

$\sin x = 0 or 2 \sin x - 1 = 0$

Solve $\sin x = 0$ on $[0, 2 π)$ :

From the unit circle, sine is zero when the terminal ray is horizontal
In $[0, 2 π)$ , this occurs at

$x = 0, x = π$

Solve $2 \sin x - 1 = 0$ on $[0, 2 π)$ :

$\sin x = \frac{1}{2}$

One solution is

$x = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$

By the symmetry of sine, the other solution in $[0, 2 π)$ is

$π - \frac{π}{6} = \frac{5 π}{6}$

Combining all the solutions gives

$x = 0, \frac{π}{6}, \frac{5 π}{6}, π$

Worked Example

Solve each of the following equations in the interval $0 \leq x < 2 π$ .

(a) $\sin x = - \frac{1}{3}$

Answer:

None of the 'special angles' has a sine equal to $- \frac{1}{3}$

so use $\sin^{- 1}$ on your calculator

$x = \sin^{- 1} (- \frac{1}{3}) = - 0.339836 . . .$

Use the symmetry of the sine function to find another solution

If $θ_{0}$ is one solution, then $π - θ_{0}$ is another solution

$x = π - (- 0.339836 . . .) = 3.481429 . . .$

$- 0.339836 . . .$ is not in the interval $0 \leq x < 2 π$

But sine has a period of $2 π$
So adding $2 π$ gives another valid solution that is in the interval

$x = - 0.339836 . . . + 2 π = 5.943348 . . .$

Round the two valid answers to 3 decimal places

$x = 3.481, 5.943 (3 d . p .)$

(b) $2 + \cos x = 1.2$

Answer:

Start by rearranging to isolate $\cos x$

$\cos x = - 0.8$

None of the 'special angles' has a cosine equal to $- 0.8$

so use $\cos^{- 1}$ on your calculator

$x = \cos^{- 1} (- 0.8) = 2.498091 . . .$

Use the symmetry of the cosine function to find another solution

If $θ_{0}$ is one solution, then $- θ_{0}$ is another solution

$x = - 2.498091 . . .$

$- 2.498091 . . .$ is not in the interval $0 \leq x < 2 π$

But cosine has a period of $2 π$
So adding $2 π$ gives another valid solution that is in the interval

$x = - 2.498091 . . . + 2 π = 3.785093 . . .$

Round the two valid answers to 3 decimal places

$x = 2.498, 3.785 (3 d . p .)$

Answer:

None of the 'special angles' has a tangent equal to $19$

so use $\tan^{- 1}$ on your calculator

$x = \tan^{- 1} (19) = 1.518213 . . .$

The tangent function has a period of $π$

So add $π$ to that to give another valid solution in the interval

$x = 1.518213 . . . + π = 4.659805 . . .$

Round the two valid answers to 3 decimal places

$x = 1.518, 4.660 (3 d . p .)$

Solving simple trigonometric inequalities

How can I solve a trigonometric inequality?

A simple trigonometric inequality has the form $\sin θ < k$ , $\cos θ \geq k$ , etc.
- The solution is an interval (or a union of intervals) within the specified solution interval
In general
- Start by solving the corresponding equation
  - i.e. with $=$ instead of the inequality sign
  - This will give you the boundary values of $θ$
- Use a graph (or the unit circle) to determine
  - which intervals between (or outside) those boundary values satisfy the inequality
- Finally, check the endpoints
  - Include them if the inequality is non-strict ( $\leq$ or $\geq$ )
  - Exclude them if the inequality is strict ( $<$ or $>$ )

How can I solve a system of trigonometric inequalities?

If a question requires more than one inequality to be satisfied at the same time:
- Find the solution set for each inequality separately
- then take the intersection
  - i.e. find the values of $θ$ that satisfy all of them
Sketching the solution intervals on a number line can help identify the intersection clearly

Worked Example

What are all values of $θ$ , $- π \leq θ \leq π$ , for which $2 \sin θ < 1$ and $2 \cos θ > - \sqrt{3}$ ?

(A) $- π \leq θ < \frac{π}{6}$

(B) $- \frac{5 π}{6} < θ < \frac{5 π}{6}$

(D) $\frac{π}{6} < θ < \frac{5 π}{6}$ only

Answer

Start by solving $2 \sin θ < 1$ on $[- π, π]$

Rewrite as $\sin θ < \frac{1}{2}$ and solve the corresponding equation
$\sin \frac{π}{6} = \frac{1}{2}$ , so

$\sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$

and by symmetry of the sine function another solution is

$θ = π - \frac{π}{6} = \frac{5 π}{6}$

Sketch the sine function on $[- π, π]$

Graph of a sine wave from -π to π with peaks at 1 and troughs at -1, intersecting the horizontal axis at -π, 0, and π, with grid lines. The dashed horizontal line y = 0.5 is shown which intersects at pi/6 and 5*pi/6.

$\sin θ < \frac{1}{2}$ to the left of $\frac{π}{6}$ and to the right of $\frac{5 π}{6}$
So the solution for $\sin θ < \frac{1}{2}$ is

$θ \in [- π, \frac{π}{6}) \cup (\frac{5 π}{6}, π]$

Now solve $2 \cos θ > - \sqrt{3}$ on $[- π, π]$

Rewrite as $\cos θ > - \frac{\sqrt{3}}{2}$ and solve the corresponding equation
$\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$ , so

$\cos θ = - \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{5 π}{6}$

and by symmetry of the cosine function another solution is

$θ = - \frac{5 π}{6}$

Sketch the cosine function on $[- π, π]$

Graph of a cosine wave from -π to π, peaking at 1, showing symmetrical rise and fall on a grid background. The dashed line at y=-sqrt(3)/2 is shown, which intersects at -5*pi/6 and 5*pi/6.

$\cos θ > - \frac{\sqrt{3}}{2}$ between these two values
So the solution for $\cos θ > - \frac{\sqrt{3}}{2}$ is

$θ \in (- \frac{5 π}{6}, \frac{5 π}{6})$

Finally, find the intersection of those two solution sets

This gives the values of $θ$ that satisfy both inequalities

$[[- π, \frac{π}{6}) \cup (\frac{5 π}{6}, π]] \cap (- \frac{5 π}{6}, \frac{5 π}{6}) = (- \frac{5 π}{6}, \frac{π}{6})$

So the solution is $- \frac{5 π}{6} < θ < \frac{π}{6}$

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Equations & Inequalities with Sinx, Cosx and Tanx (College Board AP® Precalculus): Revision Note

Solving simple trigonometric equations

What are simple trigonometric equations?

How can I find an initial solution?

How can I find other solutions from the initial solution?

What is the general process for solving a trigonometric equation in a given interval?

Examiner Tips and Tricks

How can the symmetry properties be visualized?

How are 'hidden quadratic' trigonometric equations solved?

Worked Example

Worked Example

Solving simple trigonometric inequalities

How can I solve a trigonometric inequality?

How can I solve a system of trigonometric inequalities?

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis