Chapter 7.5, Problem 48E

To determine

To find: all the real values of $x$ by solving the equation.

Answer to Problem 48E

  $\left(\frac{\pi }{6}+2\pi k\right),\left(\frac{5\pi }{6}+2\pi k\right)\text{ and }\left(\frac{\pi }{2}+2\pi k\right)$

Explanation of Solution

Given information:

  $2\sin x+\csc x=3$

Calculation:

The value of $x$ for the given equation can be evaluated as follows:

  $\begin{array}{c}2\sin x+\csc x=3\\ 2\sin x+\frac{1}{\sin x}=3\left[\text{Using quotient identity of cosecant}\right]\\ 2{\sin }^{2}x+1=3\sin x\\ 2{\sin }^{2}x-3\sin x+1=0\\ \left(2\sin x-1\right)\left(\sin x-1\right)=0\left[\text{Factors}\right]\end{array}$

Now solve the root of the above equation as follows:

  $\begin{array}{cc}\begin{array}{c}\sin x-1=0\\ \sin x=1\\ x=\frac{\pi }{2}+2\pi k\\ \left[\text{Here }k\text{ is an integer}\right]\end{array}& \begin{array}{c}2\sin x-1=0\\ \sin x=\frac{1}{2}\\ x=\frac{\pi }{6}+2\pi k\text{ and }\frac{5\pi }{6}+2\pi k\\ \left[\text{Here }k\text{ is an integer}\right]\end{array}\end{array}$

Thus the values of $x$ are $\left(\frac{\pi }{6}+2\pi k\right),\left(\frac{5\pi }{6}+2\pi k\right)\text{ and }\left(\frac{\pi }{2}+2\pi k\right)$ .