Chapter 5, Problem 52RE

To determine

To find: the all solutions of the given equation is the interval $\left[0,2\pi \right)$ and use a graphing utility to check answer.

Answer to Problem 52RE

  $x=\frac{\pi }{6},\frac{\pi }{2}$

Explanation of Solution

Given information:

Given equation

  $2{\sin }^{2}x-3\sin x=-1$ .

Calculation:

Consider the trigonometric equation

  $2{\sin }^{2}x-3\sin x=-1$

Take the constant term $-1$ on the left hand side to get

  $\begin{array}{c}2{\sin }^{2}x-3\sin x=-1\\ 2{\sin }^{2}x-3\sin x+1=0\end{array}$

Factorize by splitting the middle terms

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

Set the first factor to zero to get

  $\begin{array}{c}2\sin x-1=0\\ \sin x=\frac{1}{2}\\ x=\frac{\pi }{6}\end{array}$

Set the second factor to zero get

  $\begin{array}{c}\sin x-1=0\\ \sin x=1\\ x=\frac{\pi }{2}\end{array}$

Hence, the solutions of the equation $2{\sin }^{2}x-3\sin x=-1$ are $x=\frac{\pi }{6},\frac{\pi }{2}$ .

Now, check the result using graphing utility.

  

Hence, the solution points are $x=\frac{\pi }{6},\frac{\pi }{2}$ .