Scroll to Top

Virtual Math Learning Center

Virtual Math Learning Center Texas A&M University Virtual Math Learning Center
tags:
Factoring Sine Tangent Trigonometry Trigonometry Series

Solving Trigonometric Equations Exercise 4

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all values \(x\in [0,2\pi]\) satisfying \(\tan^2{x}+\tan^2{x}\sin{x}=0.\)

Solution: \(x=0,\pi,\frac{3\pi}{2}, 2\pi\)

Solution Method: While it might be tempting to divide everything by a \(\tan^2{x}\) here, if we do we’ll be losing a solution to the equation! (Think about \(x^2=3x\). We can clearly see that \(x=0\) is a solution to the equation. But if we divide by x, we may think \(x=3\) is the only solution.) So we cannot divide trig values out of the equation, but we can factor them out. For this equation, we have a common factor of \(\tan^2{x}\), so I can factor it out of the LHS. \[\begin{aligned} \tan^2{x}+\tan^2{x}\sin{x}&=0 \\ \tan^2{x}(1+\sin{x}) &= 0 \end{aligned}\] So I can set \(\tan^2{x}\) and \(1+\sin{x}\) equal to 0 separately.

\[\begin{aligned} \tan^2{x} &= 0 \\ \sqrt{\tan^2{x}} &= \sqrt{0} \\ \tan{x} &= 0 \\ \end{aligned}\] I know that \(\tan{x}\) is defined by the ratio identity \(\tan{x} = \frac{\sin{x}}{\cos{x}}\). So for \(\tan{x}=0\), \(\sin{x}\) must be 0. Cosine equaling zero will make tangent undefined, NOT zero! And \(\sin{x}=0\) is satisfied by the quadrantal angles \(x=0,\pi,2\pi\) (We must include both 0 and \(2\pi\) because the interval includes both as endpoints).

\[\begin{aligned} 1+\sin{x} &= 0 \\ \sin{x} &= -1 \end{aligned}\] Thinking back to the unit circle, I know that \(\sin{x} = -1\) is only satisfied in \([0,2\pi]\) by the quadrantal angle \(x=\frac{3\pi}{2}\).

So the values in \([0,2\pi]\) satisfying \(\tan^2{x}+\tan^2{x}\sin{x}=0\) are \(x=0,\pi,\frac{3\pi}{2}, 2\pi.\)

See more videos from this section

Prev Video Return to Section