Chapter 12, Problem 10CR

To determine

To solve: The equation is $2{\sin }^{2}x-\sin x-3=0,0\le x<2\pi$ .

Answer to Problem 10CR

The equation is $2{\sin }^{2}x-\sin x-3=0$ is $x=\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

The given equation is,

  $2{\sin }^{2}x-\sin x-3=0,0\le x<2\pi$

Calculation:

Consider the equation is,

  $2{\sin }^{2}x-\sin x-3=0,0\le x<2\pi \mathrm{..........}\left(1\right)$

Therefore, the limit will change as follows.

If, $x=0$ then $t=0$

And

  $x=2\pi$ then $t=0$

Therefore, the equation $\left(1\right)$ becomes as follows.

  $2t^2-t-3=0,0\le \mathrm{1..............}\left(2\right)$

So, equation $\left(2\right)$ solved,

  $\begin{array}{c}2t^2-t-3=0\\ 2t^2-3t+2t-3=0\\ t\left(2t-3\right)+1\left(2t-3\right)=0\\ \left(2t-3\right)\left(t+1\right)=0\end{array}$

  $\begin{array}{c}2t-3=0\\ t=\frac{3}{2}\end{array}$

  $\begin{array}{c}t+1=0\\ t=-1\end{array}$

Substitute the value of $t=\sin x$

  $\sin x=\frac{3}{2}$ or $\sin x=-1$

But, is not possible because sine function lie between $\left(-1,1\right)$

Therefore,

  $\begin{array}{c}x={\sin }^{-1}\left(-1\right)\\ =-\frac{\pi }{2}\\ =2\pi -\frac{\pi }{2}\\ =\frac{3\pi }{2}\end{array}$

Therefore, equation is $2{\sin }^{2}x-\sin x-3=0$ is $x=\frac{3\pi }{2}$ .