Chapter C, Problem 31E

To determine

To calculate:

The value of $x$ to satisfy the following equation:

  $\sin 2x=\cos x$

Answer to Problem 31E

The values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $\sin 2x=\cos x$ are $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\sin 2x=\cos x\\ \left[0,2\pi \right]\end{array}$

Formula Used:

The double angle theorem:

  $\sin 2x=2\sin x\cdot \cos x$

Calculation:

Know that:

  $\begin{array}{l}\sin 2x=\cos x\\ 2\sin x\cdot \cos x=\cos x\\ 2\sin x\cdot \cos x-\cos x=0\\ \cos x\left(2\sin x-1\right)=0\end{array}$

Here are two cases:

Case 1: $\cos x=0$

This case occurs at odd multiple of $\frac{\pi }{2}$ . So values of $x$ are $\frac{\pi }{2},\frac{3\pi }{2}$ for the interval $\left[0,2\pi \right]$ .

Case 1: $2\sin x-1=0$

Solve this equation:

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

So, the values of $x$ in the given interval $\left[0,2\pi \right]$ that cause sine value is equal to $\frac{1}{2}$ are $\frac{\pi }{6}$ are $\frac{5\pi }{6}$ .

Therefore, the values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $\sin 2x=\cos x$ are $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}\frac{3\pi }{2}$ .