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

Equations with Transformed Trigonometric Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Equations with transformed trigonometric functions

What is a transformed trigonometric equation?

A transformed trigonometric equation is one in which the trigonometric function is applied to a more complex argument than just $θ$
- For example
  - $\sin (2 x + \frac{π}{3}) = \frac{1}{2}, \cos (3 x - 1) = - 0.7, \tan (\frac{x}{2}) = 1$
- The argument inside the trigonometric function is itself a function of the variable
- Compare transformed sinusoidal functions like $\sin (b (θ + c))$

Equations involving transformed arguments with the reciprocal trig functions ( $\sec$ , $\csc$ , $\cot$ ) or inverse trig functions ( $\sin^{- 1}$ , $\cos^{- 1}$ , $\tan^{- 1}$ )
- can also be converted to this form

How can I solve a transformed trigonometric equation?

The most reliable approach is to
- transform the solution interval to match the new variable
- solve the simpler equation in the transformed interval
- then convert all solutions back
Start by identifying the inner expression (the argument of the trigonometric function)
- and setting it equal to a new variable $u$
  - E.g. if the equation is $\sin (2 x + \frac{π}{3}) = \frac{1}{2}$
    - then let $u = 2 x + \frac{π}{3}$
Next transform the original solution interval
- into the corresponding interval for $u$
- by applying the same operations to all parts of the inequality
  - E.g. if $\sin (2 x + \frac{π}{3}) = \frac{1}{2}$ is to be solved on the interval $0 \leq x \leq 2 π$
    - $0 \leq x \leq 2 π \Rightarrow 0 \leq 2 x \leq 4 π \Rightarrow \frac{π}{3} \leq (2 x + \frac{π}{3}) \leq \frac{13 π}{3}$
    - So $\frac{π}{3} \leq u \leq \frac{13 π}{3}$ is the transformed interval
- Note that the transformed interval is often wider (or narrower) than the original interval
  - This is important because it determines how many solutions exist
Solve the simpler equation in the transformed interval
- finding all solutions for $u$
  - E.g. $\sin u = \frac{1}{2}$ has the following solutions in $[\frac{π}{3}, \frac{13 π}{3}]$
    - $u = \frac{5 π}{6}, \frac{13 π}{6}, \frac{17 π}{6}, \frac{25 π}{6}$
Finally, convert each solution back to a value of the original variable
- by reversing the substitution
  - E.g. for $u = \frac{5 π}{6}$
    - $2 x + \frac{π}{3} = \frac{5 π}{6} \Rightarrow 2 x = \frac{π}{2} \Rightarrow x = \frac{π}{4}$
  - The full set of solutions can be found by the same method of reversing the substitution
    - $x = \frac{π}{4}, \frac{11 π}{12}, \frac{5 π}{4}, \frac{23 π}{12}$
This method is more reliable than trying to apply trigonometric identities or guess solutions directly
- especially when the interval is non-standard or the solutions are not exact angles

Examiner Tips and Tricks

The "transform the interval, solve, convert back" approach is the most reliable method for handling transformed arguments.

It works for any equation regardless of the specific values involved

A common error is to find the principal solution for the transformed variable but forget to look for additional solutions within the (often wider) transformed interval.

Another common error is forgetting to convert back to the original variable at the end.

Writing out the substitution clearly at the start helps avoid both mistakes

How do I write general solutions when no interval is specified?

If no solution interval is specified
- a trigonometric equation has infinitely many solutions
- because trigonometric functions are periodic
General solutions are written using an integer parameter (typically $n$ or $k$ ) to represent all possible solutions
- For solutions to $\sin u = p$
  - $u = u_{0} + 2 π n$ or $u = (π - u_{0}) + 2 π n$
    - where $u_{0}$ is the initial solution
    - and $n$ is any integer
- For solutions to $\cos u = p$
  - $u = u_{0} + 2 π n$ or $u = - u_{0} + 2 π n$
    - where $u_{0}$ is the initial solution
    - and $n$ is any integer
- For solutions to $\tan u = k$
  - $u = u_{0} + π n$
    - where $u_{0}$ is the initial solution
    - and $n$ is any integer
After finding the general solutions for $u$ , you can convert each one back to the original variable by reversing the substitution
- E.g. for $\sin (2 x + \frac{π}{3}) = \frac{1}{2}$ converted to $\sin u = \frac{1}{2}$
  - The initial solution is $u_{0} = \frac{π}{6}$
  - So the general solution in terms of $u$ is
    - $u = \frac{π}{6} + 2 π n$ or $u = (π - \frac{π}{6}) + 2 π n = \frac{5 π}{6} + 2 π n$
  - To convert the first part
    - $2 x + \frac{π}{3} = \frac{π}{6} + 2 π n \Rightarrow 2 x = - \frac{π}{6} + 2 π n \Rightarrow x = - \frac{π}{12} + π n$
  - To convert the second part
    - $2 x + \frac{π}{3} = \frac{5 π}{6} + 2 π n \Rightarrow 2 x = \frac{π}{2} + 2 π n \Rightarrow x = \frac{π}{4} + π n$
  - So the general solution in terms of $x$ is
    - $x = - \frac{π}{12} + π n$ or $x = \frac{π}{4} + π n$

Examiner Tips and Tricks

When a question asks for "all input values" without specifying an interval, the answer should be in general solution form using an integer parameter.

Make sure your general solution form correctly captures every solution, not just a few of them

You can use any letter you want for the integer parameter in the general solution, as long as that letter is not being used for something else in the question

E.g. if the variable in the equation is $x$ , then you would lose points on the exam if you also used $x$ as the integer parameter

How do contextual restrictions affect the solution set?

In trigonometric equations and inequalities arising from a contextual scenario
- there is often a domain restriction implied by the context
E.g. if $t$ represents time in hours after midnight
- then $t$ might be restricted to $0 \leq t < 24$
Or if $h$ represents height above the ground
- then values where $h$ is negative might not make physical sense
These contextual restrictions can limit the number of solutions to a finite set
- even when the underlying equation has infinitely many solutions

Examiner Tips and Tricks

Always check whether all the solutions you find make sense in the context of a question, and discard any that do not.

How are equations with composed inverse and trigonometric functions solved?

On the exam, some equations may involve a composition of
- an inverse trigonometric function
- with a trigonometric function
  - E.g. $\cos^{- 1} (\tan (2 x)) = 0$
The strategy is to work from the outside in
- Start by applying the inverse of the outermost function to both sides
  - E.g. $\cos^{- 1} (\tan (2 x)) = 0 \Rightarrow \cos (\cos^{- 1} (\tan (2 x))) = \cos (0) \Rightarrow \tan (2 x) = 1$
- This gets rid of the outer inverse trig function
  - and converts the equation into a simpler equation involving the inner trig function
- Then you can solve the resulting trigonometric equation using the methods described above

Worked Example

The function $f$ is given by $f (x) = \cos (1.8 x - 0.3)$ . The function $g$ is given by $g (x) = f (x) - 0.4$ . Find the zeros of $g$ on the interval $0 \leq x \leq π$ .

Answer:

Method 1

Start by setting $g (x) = 0$

$\begin{array}{rcl} g (x) & = & 0 \\ f (x) - 0.4 & = & 0 \\ \cos (1.8 x - 0.3) - 0.4 & = & 0 \end{array}$

Isolate the cosine function on one side of the equation

$\cos (1.8 x - 0.3) = 0.4$

Define the substitution

Let $u = 1.8 x - 0.3$

So the equation becomes

$\cos u = 0.4$

You also need to transform the solution interval

$0 \leq x \leq π 0 \leq 1.8 x \leq 1.8 π - 0.3 \leq 1.8 x - 0.3 \leq 1.8 π - 0.3$

$- 0.3 \leq u \leq 1.8 π - 0.3$

Solve $\cos u = 0.4$ in the transformed interval

noting that $1.8 π - 0.3 \approx 5.355$

The initial solution is

$u = \cos^{- 1} (0.4) = 1.159279 . . .$

By the symmetry of cosine, another solution is

$u = - 1.159279 . . .$

That is not in the solution interval, but adding $2 π$ to it gives another solution that is

$u = - 1.159279 + 2 π = 5.123905 . . .$

Both $u \approx 1.159$ and $u \approx 5.124$ are within the transformed interval

Convert the solutions back to $x$

$u = 1.8 x - 0.3 \Rightarrow 1.8 x = u + 0.3 \Rightarrow x = \frac{u + 0.3}{1.8}$

Therefore

$x_{1} = \frac{1.159279 . . . + 0.3}{1.8} = 0.810710 . . .$

$x_{2} = \frac{5.123905 . . . + 0.3}{1.8} = 3.013281 . . .$

Round the answers to 3 decimal places

$x = 0.811, 3.013 (3 d . p .)$

Method 2

If you do not need to show your working to earn full marks (e.g. if a question like this were presented instead as a multiple choice question), you can also solve this sort of equation using your graphing calculator

Graph the function $g (x) = \cos (1.8 x - 0.3) - 0.4$
And identify the $x$ -axis crossings in the interval $0 \leq x \leq π$

Graph of a transformed cosine wave with points marked at (0.81071,0) and (3.01328,0) on a grid, showing peaks and troughs.

Round the answers to 3 decimal places

$x = 0.811, 3.013 (3 d . p .)$

Worked Example

The function $h$ is given by $h (x) = 2 \cos (3 x) + 1$ . Find all input values in the domain of $h$ that yield an output value of $0$ .

Answer:

Set $h (x) = 0$ and rearrange to isolate the cosine function

$2 \cos (3 x) + 1 = 0$

$\cos (3 x) = - \frac{1}{2}$

Define the substitution

Let $u = 3 x$

So the equation becomes

$\cos u = - \frac{1}{2}$

Since no interval is specified, you need to find the general solution for $u$

Find the initial solution

$u = \cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$

And by the symmetry of the cosine function, another solution is

$u = - \frac{2 π}{3}$

Cosine has a period of $2 π$

so adding integer multiples of $2 π$ to those two solutions gives all the other solutions

$u = \frac{2 π}{3} + 2 π n or u = - \frac{2 π}{3} + 2 π n$

Finally, $u = 3 x \Rightarrow x = \frac{u}{3}$

So you can convert the solution back to $x$ by dividing each expression by $3$

$x = \frac{2 π}{9} + (\frac{2 π}{3}) n or x = - \frac{2 π}{9} + (\frac{2 π}{3}) n$

Worked Example

The function $m$ is given by $m (x) = \sin^{- 1} (\cos (2 x))$ . Find all input values in the domain of $m$ that yield an output value of $0$ .

Answer:

Set $m (x) = 0$

$\sin^{- 1} (\cos (2 x)) = 0$

Apply the sine function to both sides

$\sin (\sin^{- 1} (\cos (2 x))) = \sin (0)$

$\cos (2 x) = 0$

Now solve $\cos (2 x) = 0$ for all real $x$

$\sin^{- 1}$ has a domain of $[- 1, 1]$
- But $\cos 2 x$ only outputs values in $[- 1, 1]$
- So any real number is a valid input for $m (x)$
I.e. the domain for $m (x)$ is all real numbers

Define the substitution

Let $u = 2 x$

So the equation becomes

$\cos u = 0$

The domain is all real numbers, so you need to find the general solution for $u$

Find the initial solution

$u = \cos^{- 1} (0) = \frac{π}{2}$

And by the symmetry of the cosine function, another solution is

$u = - \frac{π}{2}$

Cosine has a period of $2 π$

so adding integer multiples of $2 π$ to those two solutions gives all the other solutions

$u = \frac{π}{2} + 2 π n or u = - \frac{π}{2} + 2 π n$

Finally, $u = 2 x \Rightarrow x = \frac{u}{2}$

So you can convert the solution back to $x$ by dividing each expression by $2$

$x = \frac{π}{4} + π n or x = - \frac{π}{4} + π n$

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Equations with Transformed Trigonometric Functions (College Board AP® Precalculus): Revision Note

Equations with transformed trigonometric functions

What is a transformed trigonometric equation?

How can I solve a transformed trigonometric equation?

Examiner Tips and Tricks

How do I write general solutions when no interval is specified?

Examiner Tips and Tricks

How do contextual restrictions affect the solution set?

Examiner Tips and Tricks

How are equations with composed inverse and trigonometric functions solved?

Worked Example

Worked Example

Worked Example

Unlock more, it's free!

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

Author: Roger B

Reviewer: Mark Curtis