Chapter 5, Problem 14CT

To determine

To find: the all solutions of the equation in the interval $\left[0,2\pi \right)$ .

Answer to Problem 14CT

  $\alpha =\frac{\pi }{2},\frac{\pi }{6},\frac{3\pi }{2},\frac{5\pi }{6}$ .

Explanation of Solution

Given information:

Given function is

  $\sin 2\alpha -\cos \alpha =0$ .

Calculation:

Consider the expression

  $\sin 2\alpha -\cos \alpha =0$

Angle lies in the interval $\left(0,2\pi \right)$ .

Use following trigonometric formula,

  $\sin 2\theta =2\sin \theta \cos \theta$

Rewrite the expression as,

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

It implies the following,

  $\cos \alpha =0$ and $\sin \alpha =\frac{1}{2}$

Now,

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

Similarly,

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

Therefore, the values of $\alpha$ are $\alpha =\frac{\pi }{2},\frac{\pi }{6},\frac{3\pi }{2},\frac{5\pi }{6}$ .