Chapter 5.4, Problem 7E

Solution

To find: The solutions of the given equation in the interval $\left(0,2\pi \right)$ .

The solutions of the given equation are $\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$ .

Given information:

The given equation is $\cos 2x=\sin x$ .

Formula used:

The double angle identity for $\cos 2x$ is given by $\cos 2x=1-2{\sin }^{2}x$ .

Calculation:

Consider the given equation.

  $\cos 2x=\sin x$

Apply the double angle identity in the given equation.

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

Consider $\sin x=u$ . Substitute $\sin x$ for $u$ in the above equation and simplify.

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

Substitute $\sin x$ for $u$ in the above equation.

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

Apply the zero-factor principle and equate the first factor to zero.

  $\begin{array}{c}\sin x+1=0\\ \sin x=-1\\ x=\frac{3\pi }{2}\left[\text{Subject to interval }\left(0,2\pi \right)\right]\end{array}$

Equate the second factor to zero.

  $\begin{array}{c}2\sin x-1=0\\ \sin x=\frac{1}{2}\\ x=\frac{\pi }{6},\frac{5\pi }{6}\left[\text{Subject to interval }\left(0,2\pi \right)\right]\end{array}$

Therefore, the solutions of the given equation are $\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$ .