# How do you use inverse trigonometric functions to find the solutions of the equation that are in the interval 0 2π)?

Dec 8, 2014

To find solutions to a trigonometric equation start by taking the inverse trig function (like inverse sin, inverse cosine, inverse tangent) of both sides of the equation and then set up reference angles to find the rest of the answers.

This inverse method gives the one answer on the interval for which each inverse trigonometric function is defined:

For ${\sin}^{-} 1 \to \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

For ${\cos}^{-} 1 \to \left[0 , \pi\right]$

For ${\tan}^{-} 1 \to \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

To find the rest of the answers (answers often come in pairs unless a max or min) using reference triangles, draw your reference triangle on the set of axes in the proper quadrant/location. Remember negative angles go in quadrant IV.

For the other answer to a sine equation, reflect the reference angle over the y-axis. (Remember that sine is the y-coordinate of the unit circle and you need to find the angle with the matching y-coordinate.) A short cut is to subtract your answer from $\pi$.

For the other answer to a cosine equation, reflect the reference angle over the x-axis. (Remember that the cosine is the x-coordinate on the unit circle and you need to find the angle with the matching x-coordinate. A short cut is to subtract your answer from $2 \pi$.

In the case of sine if your first angle is negative, add $2 \pi$ to that angle to get the positive version. In the case of tangent, if your first angle is negative, add $\pi$ to get your first answer and add $\pi$ again to get your second answer.
Remember exact answers can be found from famous ratios $\frac{1}{2} , \frac{\sqrt{3}}{2} , \frac{\sqrt{2}}{2} , 0 , 1$ on the unit circle, and decimal approximations can be found using your calculator set in the proper mode.