# How to solve 4sin^2((3x+pi)/6)=3?

## So, $\sin \left(\frac{3 x + \pi}{6}\right) = \pm \frac{\sqrt{3}}{2}$. Then what?

Jun 7, 2018

$x = \left(2 k + 1\right) \pi$v$\left(2 k - \frac{1}{3}\right) \pi$

#### Explanation:

We have $\sin \left(2 \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ (see https://socratic.org/questions/what-is-sin-2pi-3-equal-to)

Therefore
$\sin \left(\frac{3 x + \pi}{6}\right) = \pm \sin \left(2 \frac{\pi}{3}\right)$
$\frac{3 x + \pi}{6} = \left(\frac{2}{3} \pi + 2 k \pi\right)$ v $\left(- \frac{2}{3} \pi + 2 k \pi\right)$
$- \frac{2}{3} \pi$ should have the same sin value as $\frac{4}{3} \pi$
so that we get
$3 x + \pi = 6 \left(\frac{2}{3} \pi + 2 k \pi\right)$ v $6 \left(\frac{4}{3} \pi + 2 k \pi\right)$
$3 x = \left(4 \pi - \pi + 12 k \pi\right)$v$\left(12 \pi - \pi + 12 k \pi\right)$
$3 x = \left(3 \pi + 12 k \pi\right)$v$\left(11 \pi + 12 k \pi\right)$
$x = \left(\pi + 4 k \pi\right)$ v $\left(\frac{11}{3} \pi + 4 k \pi\right)$
$x = \left(2 k + 1\right) \pi$ v $\left(2 k - \frac{1}{3}\right) \pi$

Jun 8, 2018

$x = \left(2 k + 1\right) \frac{\pi}{3}$

#### Explanation:

$4 {\sin}^{2} \left(\frac{3 x + \pi}{6}\right) = 3$
$\sin \left(\frac{3 x + \pi}{6}\right) = \pm \frac{\sqrt{3}}{2}$
a. $\sin \left(\frac{3 x + \pi}{6}\right) = \frac{\sqrt{3}}{2}$
Trig table and unit circle give 2 solutions for $\frac{3 x + \pi}{6}$:
1. $\frac{3 x + \pi}{6} = \frac{\pi}{3}$ --> $3 x + \pi = 2 \pi$
$3 x = 2 \pi - \pi = \pi + 2 k \pi$ -->$x = \frac{\pi}{3} + \frac{2 k \pi}{3}$
2. $\frac{3 x + \pi}{6} = \frac{2 \pi}{3}$ --> $3 x + \pi = 4 \pi$
$3 x = 3 \pi + 2 k \pi$ --> $x = \pi + \frac{2 k \pi}{3}$
General answer: $x = \left(2 k + 1\right) \frac{\pi}{3}$
b. $\sin \left(\frac{3 x + \pi}{6}\right) = - \frac{\sqrt{3}}{2}$
Trig table and unit circle give 2 solutions for $\frac{3 x + \pi}{6}$
1. $\frac{3 x + \pi}{6} = \left(4 \frac{\pi}{3}\right)$ --> $\left(3 x + \pi\right) = 8 \pi$
$3 x = 7 \pi + 2 k \pi$ --> $x = \frac{7 \pi}{3} + \frac{2 k \pi}{3}$, or
$x = \frac{\pi}{3} + \frac{2 k \pi}{3}$
2. $\frac{3 x + \pi}{6} = \frac{5 \pi}{3}$ --> $3 x + \pi = 10 \pi$
$3 x = 9 \pi + 2 k \pi$ --> $x = 3 \pi + \frac{2 k \pi}{3}$, or
x = pi + (2kpi)/3
General answer: $x = \left(2 k + 1\right) \frac{\pi}{3}$