# Solve equation: (Is my answer correct ?)

## ${\sin}^{2} x + {\sin}^{2} 2 x = {\sin}^{2} 3 x$ ${\sin}^{2} x + {\left(2 \sin x \cos x\right)}^{2} = \left[\sin x \left(3 - 4 {\sin}^{2} x\right)\right]$ ${\sin}^{2} x + 4 {\sin}^{2} x {\cos}^{2} x = {\sin}^{2} x \left(9 - 24 {\sin}^{2} x + 16 {\sin}^{4} x\right)$ ${\sin}^{2} x + 4 {\sin}^{2} x {\cos}^{2} x = 16 {\sin}^{6} x - 24 {\sin}^{4} x + 9 {\sin}^{2} x$ $4 {\sin}^{2} x \left(4 {\sin}^{4} x - 6 {\sin}^{2} x + 2 - {\cos}^{2} x\right) = 0$ ${\sin}^{2} x = 0$ x=kπ ⋁ $4 {\sin}^{4} x - 6 {\sin}^{2} x + 2 - 1 + {\sin}^{2} x = 0$ $4 {\sin}^{4} x - 5 {\sin}^{2} x + 1 = 0$ ${\sin}^{2} x = t$ $4 {t}^{2} - 5 t + 1 = 0$ ∆=9 ${t}_{1} = \frac{1}{4}$ and ${t}_{2} = 1$ so... for ${t}_{1} = \frac{1}{4}$ : $\frac{1}{2} = \sin x$ or $- \frac{1}{2} = \sin x$ x=π/6 + 2kπ or x=-π/6 + 2kπ for ${t}_{2} = 1$ : $\sin x = 1 \mathmr{and} \sin x = - 1$ x=π/2+2kπ or x=-π/2+2kπ   x=π/6 + kπ ⋁ x=π/2 + kπ

Jun 27, 2018

$x = \pm {37}^{\circ} 76 + k {180}^{\circ}$
$x = k \pi$, and $x = \frac{\pi}{2} + k \pi$

#### Explanation:

${\sin}^{2} x - {\sin}^{2} 3 x + {\sin}^{2} 2 x = 0$
$\left(1 - \cos 2 x\right) - \left(1 - \cos 6 x\right) + {\sin}^{2} 2 x = 0$
$\left(\cos 6 x - \cos 2 x\right) + {\sin}^{2} 2 x = 0$
$- 2 \sin 4 x . \sin 2 x + {\sin}^{2} 2 x = 0$
$\sin 2 x \left(\sin 2 x - 2 \sin 4 x\right) = \sin 2 x \left(\sin 2 x - 4 \sin 2 x . \cos 2 x\right) = 0$
${\sin}^{2} 2 x \left(1 - 4 \cos 2 x\right) = 0$
Either factor should be zero
.
a. $\sin 2 x = 0$. Unit circle gives:
1. $2 x = 2 k \pi$ --> $x = k \pi$
2. $2 x = \pi + 2 k \pi$ --> $x = \frac{\pi}{2} + k \pi$
b. $4 \cos 2 x = 1$ --> $\cos 2 x = \frac{1}{4}$
Calculator and unit circle give -->
1. $2 x = \pm {75}^{\circ} 52 + k {360}^{\circ}$ -->
$x = \pm {37}^{\circ} 76 + k {180}^{\circ}$