Chapter 12, Problem 10CR

To determine

The solution of the equation $2{\sin }^{2}x-\sin x-3=0,0\le x<2\pi$.

Answer to Problem 10CR

Solution:

The solution set of the equation $2{\sin }^{2}x-\sin x-3=0,0\le x<2\pi$, is $\left\{\frac{3\pi }{2}\right\}$.

Explanation of Solution

Given information:

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

Explanation:

Consider equation $2{\sin }^{2}x-\sin x-3=0$.

Factoring,

$\Rightarrow \left(2\sin x-3\right)\left(\sin x+1\right)=0$.

$\Rightarrow 2\sin x-3=0$ or $\sin x+1=0$.

$2\sin x-3=0\Rightarrow \sin x=\frac{3}{2}$, which is impossible.

$\sin x+1=0\Rightarrow \sin x=-1$

$\Rightarrow x=\frac{3\pi }{2}$ as $0\le x<2\pi$

Hence, the solution set of equation $2{\sin }^{2}x-\sin x-3=0,0\le x<2\pi$, is $\left\{\frac{3\pi }{2}\right\}$.