# How do you solve sin2x = √2 cosx, 0≤x≤ π?

## sin2x = $\sqrt{2}$ cosx, 0$\le$x$\le$$\pi$?

Mar 24, 2017

pi/2; pi/4; (3pi)/4

#### Explanation:

Use trig identity : sin 2x = 2sin x.cos x
Substitute in the equation sin 2x by 2sin x.cos x
$2 \sin x . \cos x - \sqrt{2} \cos x = 0$
$\cos x \left(2 \sin x - \sqrt{2}\right) = 0$
Either one of the 2 factors should be zero:
a. cos x = 0 --> $x = \frac{\pi}{2}$, and $x = \frac{3 \pi}{2}$
b. $2 \sin x - \sqrt{2} = 0$
$\sin x = \frac{\sqrt{2}}{2}$
Trig unit circle gives 2 solutions:
$x = \frac{\pi}{4} \mathmr{and} x = \frac{3 \pi}{4}$
Answers for $\left(0 , \pi\right)$
pi/2; pi/4; (3pi)/4