Question #db34a

May 25, 2017

$x = \frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6}$ and $x = \frac{3 \pi}{2}$

Explanation:

$\cos 2 x = \sin x$

Substitute using the trig identity $\cos 2 x = 1 - 2 {\sin}^{2} x$

$1 - 2 {\sin}^{2} x = \sin x$

Subtract $1 - 2 {\sin}^{2} x$ from both sides.

$0 = 2 {\sin}^{2} x + \sin x - 1$

Factor

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

Set each factor equal to zero and solve.

$2 \sin x - 1 = 0$ and $\sin x + 1 = 0$

$\sin x = \frac{1}{2}$ and $\sin x = - 1$

From the unit circle for $0 \le x < 2 \pi$

$x = \frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6}$ and $x = \frac{3 \pi}{2}$

May 25, 2017

The solutions are $S = \left\{\frac{\pi}{6} + 2 k \pi , \frac{5}{6} \pi + 2 k \pi , \frac{3}{2} \pi + 2 k \pi\right\}$, $k \in \mathbb{Z}$

Explanation:

We need

$\cos 2 x = 1 - 2 {\sin}^{2} x$

Therefore,

our equation becomes

$\cos 2 x = \sin x$

$1 - 2 {\sin}^{2} x = \sin x$

$2 {\sin}^{2} x + \sin x - 1 = 0$

We solve this like a quadratic equation

The discriminant is

$\Delta = {b}^{2} - 4 a c = 1 - 4 \cdot \left(2\right) \cdot \left(- 1\right) = 1 + 8 = 9$

As,

$\Delta > 0$, there are 2 real solutions

$\sin x = \frac{- b + \sqrt{\Delta}}{2 a} = \frac{- 1 + \sqrt{9}}{2 \cdot 2} = \frac{2}{4} = \frac{1}{2}$

$x = \frac{\pi}{6} + 2 k \pi$ and $x = \frac{5}{6} \pi + 2 k \pi$, $k \in \mathbb{Z}$

and

$\sin x = \frac{- b - \sqrt{\Delta}}{2 a} = \frac{- 1 - \sqrt{9}}{2 \cdot 2} = - \frac{4}{4} = - 1$

$x = \frac{3}{2} \pi + 2 k \pi$, $k \in \mathbb{Z}$