# How do you solve 4\sin ^ { 2} 3x - 1= 0?

Nov 24, 2017

$x = \frac{\pm \pi}{18} + \frac{2 k \pi}{3} \mathmr{and} \frac{\pm 5 \pi}{18} + \frac{2 k \pi}{3} , k \in \mathbb{N}$

#### Explanation:

4sin^2 3x−1=0

This is an equation of the type

$4 {y}^{2} - 1 = 0$

So we must start from that:

$4 {y}^{2} - 1 = 0$

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

So, $y = \pm \frac{1}{2}$

Remember $\to y = \sin 3 x$

$\sin 3 x = \pm \frac{1}{2}$

See in the trigonometrical circle and realize that:

$3 x = \pm \frac{\pi}{6} + 2 k \pi \mathmr{and} \frac{\pm 5 \pi}{6} + 2 k \pi , k \in \mathbb{N}$

$x = \frac{\pm \pi}{18} + \frac{2 k \pi}{3} \mathmr{and} \frac{\pm 5 \pi}{18} + \frac{2 k \pi}{3} , k \in \mathbb{N}$