AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsTrigonometric Equations, Inequalities & IdentitiesEquations & Inequalities with Inverse & Reciprocal Trigonometric Functions

Equations & Inequalities with Inverse & Reciprocal Trigonometric Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Equations & inequalities with arcsin, arccos and arctan

How can I solve equations involving inverse trigonometric functions?

An equation of the form $\sin^{- 1} (expression) = c$ (or similar with $\cos^{- 1}$ or $\tan^{- 1}$ ) can be solved by applying the corresponding trigonometric function to both sides
- For example, given $\sin^{- 1} (x) = c$ , applying sine to both sides gives:
  - $\sin (\sin^{- 1} (x)) = \sin c \Rightarrow x = \sin c$
- This converts the inverse trig equation into a simple trig equation

In general:
- Start by applying the appropriate trigonometric function to both sides to remove the inverse trig function
- Then solve the resulting equation
- Check that the solution is in the domain of the original inverse trig function
  - i.e. that the solution is a valid input for the inverse trig function
    - in the interval $[- 1, 1]$ for $\sin^{- 1}$ and $\cos^{- 1}$
    - or any real number for $\tan^{- 1}$
- Also check that the value of $c$ on the right-hand side of the original equation is in the range of the inverse trig function
  - i.e. $[- \frac{π}{2}, \frac{π}{2}]$ for $\sin^{- 1}$ , $[0, π]$ for $\cos^{- 1}$ , $(- \frac{π}{2}, \frac{π}{2})$ for $\tan^{- 1}$
  - If $c$ is outside this range, the equation has no solution

Worked Example

Solve each of the following equations.

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

Answer:

Apply the sine function to both sides

$\begin{array}{rcl} \sin (\sin^{- 1} (2 x - 1)) & = & \sin \frac{π}{6} \\ 2 x - 1 & = & \sin \frac{π}{6} \end{array}$

Substitute in the value of $\sin \frac{π}{6}$ and solve for $x$

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

$2 x = \frac{3}{2}$

$x = \frac{3}{4}$

Check the domain

$\sin^{- 1}$ requires
- $- 1 \leq 2 x - 1 \leq 1 \Rightarrow 0 \leq 2 x \leq 2 \Rightarrow 0 \leq x \leq 1$
Since $\frac{3}{4}$ is in $[0, 1]$ , the solution is valid.

Check the range

$\frac{π}{6}$ is in $[- \frac{π}{2}, \frac{π}{2}]$ , the range of $\sin^{- 1}$ ,
so the original equation has a solution

$x = \frac{3}{4}$

(b) $\cos^{- 1} (3 x) = \frac{5 π}{6}$

Answer:

Apply the cosine function to both sides

$\cos (\cos^{- 1} (3 x)) = \cos \frac{5 π}{6}$

$3 x = \cos \frac{5 π}{6}$

Substitute in the value of $\cos \frac{5 π}{6}$ and solve for $x$

$3 x = - \frac{\sqrt{3}}{2}$

$x = - \frac{\sqrt{3}}{6}$

Check the domain

$\cos^{- 1}$ requires
- $- 1 \leq 3 x \leq 1 \Rightarrow - \frac{1}{3} \leq x \leq \frac{1}{3}$
Since $- \frac{1}{3} = - \frac{2}{6} = - \frac{\sqrt{4}}{6} < - \frac{\sqrt{3}}{6}$
- $- \frac{\sqrt{3}}{6}$ is in $[- \frac{1}{3}, \frac{1}{3}]$ and the solution is valid.

Check the range

$\frac{5 π}{6}$ is in $[0, π]$ , the range of $\cos^{- 1}$
so the original equation has a solution.

$x = - \frac{\sqrt{3}}{6}$

How can I solve inequalities involving inverse trigonometric functions?

Note these important behaviors of the three inverse trigonometric functions:
- $\sin^{- 1}$ and $\tan^{- 1}$ are always increasing everywhere on their domains
- $\cos^{- 1}$ is always decreasing everywhere on its domain
This makes inverse trig inequalities relatively straightforward to solve
- Start by solving the corresponding equation (replacing the inequality sign with $=$ )
  - This will give you the boundary value
- Then use the behavior of the inverse trig function to determine which side of the boundary satisfies the inequality
- Check the domain of the inverse trig function
  - for $\sin^{- 1}$ and $\cos^{- 1}$ , the expression inside must lie in $[- 1, 1]$
  - for $\tan^{- 1}$ , any real number is allowed
- Combine the inequality with the domain restriction to give the final solution
E.g. to solve $\sin^{- 1} (2 x - 1) > \frac{π}{6}$
- The corresponding equation is $\sin^{- 1} (2 x - 1) = \frac{π}{6}$
  - $\begin{array}{rcl} 2 x - 1 & = & \sin \frac{π}{6} \Rightarrow 2 x - 1 = \frac{1}{2} \Rightarrow x = \frac{3}{4} \end{array}$
- Since $\sin^{- 1}$ is increasing, $\sin^{- 1} (2 x - 1)$ will be greater than $\frac{π}{6}$ to the right of $x = \frac{3}{4}$
  - This gives $x > \frac{3}{4}$
- But the domain restriction is
  - $- 1 \leq 2 x - 1 \leq 1 \Rightarrow 0 \leq x \leq 1$
- Combining these gives the final solution
  - $\frac{3}{4} < x \leq 1$

Equations & inequalities with sec, csc and cot

How can I solve reciprocal trigonometric equations?

A simple reciprocal trigonometric equation is one of the form
- $\sec θ = k, \csc θ = k, or \cot θ = k$
  - where $k$ is a constant

More generally, a question may give an equation that simplifies to one of these forms after some algebraic rearrangement
- e.g. $3 \sec θ + 1 = 7$ , which can be rearranged to give $\sec θ = 2$
The goal is to find all values of $θ$ in a specified solution interval that make the equation true
The most reliable approach is to convert the equation into an equation involving sine, cosine, or tangent
- by using the reciprocal relationships
  - $\sec θ = \frac{1}{\cos θ}, \csc θ = \frac{1}{\sin θ}, \cot θ = \frac{1}{\tan θ}$
- Once converted, the equation can be solved using the methods for simple trigonometric equations

In general:
- Start by rearranging the equation to isolate the reciprocal trig function on one side
  - i.e. in the form $\sec θ = k$ , $\csc θ = k$ , or $\cot θ = k$
- Then take the reciprocal of both sides to convert to an equation involving cosine, sine, or tangent
  - $\sec θ = k$ becomes $\cos θ = \frac{1}{k}$
  - $\csc θ = k$ becomes $\sin θ = \frac{1}{k}$
  - $\cot θ = k$ becomes $\tan θ = \frac{1}{k}$
- Solve the resulting equation using the methods for simple trigonometric equations
  - i.e. find an initial solution, then use symmetry and periodicity to find all other solutions in the interval

When does a reciprocal trigonometric equation have no solution?

The secant and cosecant functions have a range of $(- \infty, - 1] \cup [1, \infty)$
- i.e. they never take values in the interval $(- 1, 1)$
Therefore, equations like $\sec θ = 0.5$ or $\csc θ = - 0.3$ have no solution
- because the right-hand side is not in the range of the function
This can also be seen from the reciprocal step
- E.g. $\sec θ = 0.5$ would give $\cos θ = \frac{1}{0.5} = 2$
  - but cosine cannot exceed $1$
The cotangent function outputs all real number values
- so $\cot θ = k$ always has solutions

Examiner Tips and Tricks

When solving an equation involving secant, cosecant, or cotangent, always check whether the right-hand side (after isolating the reciprocal function) is actually in the range of that function.

If you are asked to solve $\sec θ = k$ and find that $| k | < 1$ , the equation has no solution and you can stop without doing any further work.

Recognising this quickly can save valuable time on the exam.

Worked Example

Let $f (x) = 3 - 2 \csc x$ and $g (x) = 7$ . In the $x y$ -plane, what are the $x$ -coordinates of the points of intersection of the graphs of $f$ and $g$ for $0 \leq x < 2 π$ ?

(A) $x = \frac{π}{6}$ and $x = \frac{5 π}{6}$

(B) $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$

(D) $x = \frac{7 π}{6}$ and $x = \frac{11 π}{6}$

Answer

The graphs of $f$ and $g$ intersect where $f (x) = g (x)$ :

$3 - 2 \csc x = 7$

Isolate the cosecant term

$- 2 \csc x = 4$

$\csc x = - 2$

Take the reciprocal of both sides to convert to a sine equation

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

Find an initial solution

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

Use the symmetry of the sine function to find another solution

$x = π - (- \frac{π}{6}) = \frac{7 π}{6}$

$- \frac{π}{6}$ is not in the interval $0 \leq x < 2 π$

But adding $2 π$ to it gives another valid solution that is in the interval

$x = - \frac{π}{6} + 2 π = \frac{11 π}{6}$

So the correct answer is

(D) $x = \frac{7 π}{6}$ and $x = \frac{11 π}{6}$

How can I solve inequalities with reciprocal trigonometric functions?

A reciprocal trigonometric inequality has the form $\sec θ < k$ , $\csc θ \geq k$ , $\cot θ > k$ , etc.
The general approach is similar to solving the equations
- Start by rearranging to isolate the reciprocal trig function
- Solve the corresponding equation (with $=$ instead of the inequality sign)
  - This will give you the boundary values
- Identify which intervals satisfy the inequality
  - E.g. by using a graph or by testing values from each interval
Take particular care with reciprocal trig inequalities because the reciprocal functions have vertical asymptotes
- These split the solution intervals
- E.g. when solving an inequality involving secant
  - the asymptotes at $θ = \frac{π}{2} + k π$ may divide the solution set into multiple disjoint intervals

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Equations & Inequalities with Inverse & Reciprocal Trigonometric Functions (College Board AP® Precalculus): Revision Note

Equations & inequalities with arcsin, arccos and arctan

How can I solve equations involving inverse trigonometric functions?

Worked Example

How can I solve inequalities involving inverse trigonometric functions?

Equations & inequalities with sec, csc and cot

How can I solve reciprocal trigonometric equations?

When does a reciprocal trigonometric equation have no solution?

Examiner Tips and Tricks

Worked Example

How can I solve inequalities with reciprocal trigonometric functions?

Unlock more, it's free!

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

Author: Roger B

Reviewer: Mark Curtis