Chapter 5, Problem 19CT

To determine

To calculate: The solution of the expression $\sin 2\alpha -\cos \alpha =0$ in the interval $[0,2\pi )$ .

Answer to Problem 19CT

The solutions are $\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}$ .

Explanation of Solution

Given Information:

The trigonometric expression is $\sin 2\alpha -\cos \alpha =0$ .

Range of $\alpha$ is $[0,2\pi )$ .

Formula Used

The trigonometric formula $\cos \left(2A\right)=2\sin A\cos B$ .

Calculation:

The given expression is $\sin 2\alpha -\cos \alpha =0$ .

Now, use the trigonometric formula $\sin \left(2A\right)=2\sin A\cos A$ .

  $\begin{array}{c}\sin 2\alpha -\cos \alpha =0\\ 2\sin \alpha \cos \alpha -\cos \alpha =0\\ \cos \alpha \left(2\sin \alpha -1\right)=0\end{array}$

So, $\cos \alpha =0$ or $2\sin \alpha -1=0$ .

For $\cos \alpha =0$ .

  $\begin{array}{c}\cos \alpha =0\\ \alpha ={\cos }^{-1}0\\ \alpha =\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

And for $2\sin \alpha -1=0$ .

  $\begin{array}{c}2\sin \alpha -1=0\\ \sin \alpha =\frac{1}{2}\\ \alpha ={\sin }^{-1}\frac{1}{2}\\ \alpha =\frac{\pi }{6},\frac{5\pi }{6}\end{array}$

Hence, the solutions are $\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}$ .