Chapter 5.1, Problem 54E

Solution

To find: All solutions to the equation $\sin x{\tan }^{2}x=\sin x$ in the interval $\left[0,2\pi \right)$ .

The solutions to the equation $\sin x{\tan }^{2}x=\sin x$ in the interval $\left[0,2\pi \right)$ is $\left\{0,\pi ,\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right\}$ .

Given information:

The given trigonometric equation is $\sin x{\tan }^{2}x=\sin x$ .

Formula used:

The general solution of $\sin x=0$ in the interval $\left[0,2\pi \right)$ is $\left\{0,\pi \right\}$ .

The general solution of $\tan x=1$ in the interval $\left[0,2\pi \right)$ is given by $\left\{\frac{\pi }{4},\frac{5\pi }{4}\right\}$ and for $\tan x=-1$ in the interval $\left[0,2\pi \right)$ is given by $\left\{\frac{3\pi }{4},\frac{7\pi }{4}\right\}$ .

Calculation:

Consider the given equation

  $\begin{array}{c}\sin x{\tan }^{2}x=\sin x\\ \sin x{\tan }^{2}x-\sin x=0\end{array}$

Take the common factor out.

  $\sin x\left({\tan }^{2}x-1\right)=0$

On simplify, there are two possibilities.

Either $\sin x=0$

Or $\begin{array}{c}{\tan }^{2}x=1\\ \tan x=\pm 1\end{array}$

For $\sin x=0$ , the general solution in the interval $\left[0,2\pi \right)$ includes,

  $x=0,\pi$

For $\tan x=\pm 1$ , the general solution in the interval $\left[0,2\pi \right)$ includes,

  $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

Thus, the solutions to the equation $\sin x{\tan }^{2}x=\sin x$ in the interval $\left[0,2\pi \right)$ is $\left\{0,\pi ,\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right\}$ .