Chapter 14.8, Problem 2OE

To determine

To find:the number of solutions does the given equation have the interval $0\le x\le 2\pi$ .

Answer to Problem 2OE

Two solution, $x=-\frac{\pi }{2},\frac{3\pi }{2}$ .

Explanation of Solution

Given:

The given equation is $\sin x=-1$ .

Calculation:

As thegiven equation is $\sin x=-1$ .

Now, to find the solutions of the given equation. Therefore,

  $\begin{array}{l}\sin x=-1\{\text{given equation}}\\ \therefore x={\sin }^{-1}\left(-1\right)\\ x=-{\sin }^{-1}\left(1\right)\{{\sin }^{-1}\left(-x\right)=-{\sin }^{-1}x\}\\ x=-{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)\left\{\left(\sin \frac{\pi }{2}\right)=1\right\}\\ x=-\frac{\pi }{2}\end{array}$

Now, to find the other solution of the equation.

As $\sin \theta$ is positive in quadrant II, therefore,

  $\begin{array}{l}x=\pi -\theta \\ x=\pi -\left(-\frac{\pi }{2}\right)=\pi +\frac{\pi }{2}=\frac{3\pi }{2}\end{array}$

Hence, the solutions have two solution i.e. $x=-\frac{\pi }{2},\frac{3\pi }{2}$ .