How do you find all sin^2 3x-sin^2x=0 in the interval [0,2pi)?

Jul 15, 2017

The solutions are $S = \left\{0 , \frac{1}{4} \pi , \frac{1}{2} \pi , \frac{3}{4} \pi , \pi , \frac{5}{4} \pi , \frac{3}{2} \pi , \frac{7}{4} \pi\right\}$

Explanation:

We need

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

$\sin 2 x = 2 \sin x c \otimes$

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

We start by calculating $\sin 3 x$

$\sin 3 x = \sin \left(2 x + x\right)$

$= \sin 2 x \cos x + \cos 2 x \sin x$

$= 2 \sin x {\cos}^{2} x + \sin x \left(1 - 2 {\sin}^{2} x\right)$

$= 2 \sin x \left(1 - {\sin}^{2} x\right) + \sin x - 2 {\sin}^{3} x$

$= 2 \sin x - 2 {\sin}^{3} x + \sin x - 2 {\sin}^{3} x$

$= 3 \sin x - 4 {\sin}^{3} x$

Therefore,

${\sin}^{2} \left(3 x\right) - {\sin}^{2} x = 0$

${\left(3 \sin x - 4 {\sin}^{3} x\right)}^{2} - {\sin}^{2} x = 0$

$9 {\sin}^{2} x - 24 {\sin}^{4} x + 16 {\sin}^{6} x - {\sin}^{2} x = 0$

$16 {\sin}^{6} x - 24 {\sin}^{4} x + 8 {\sin}^{2} x = 0$

$8 {\sin}^{2} x \left(2 {\sin}^{4} x - 3 {\sin}^{2} x + 1\right) = 0$

$8 {\sin}^{2} x \left(2 {\sin}^{2} x - 1\right) \left({\sin}^{2} x - 1\right) = 0$

$8 {\sin}^{2} x \left(\sqrt{2} \sin x + 1\right) \left(\sqrt{2} \sin x - 1\right) \left(\sin x + 1\right) \left(\sin x - 1\right) = 0$

So,

${\sin}^{2} x = 0$, $\implies$, $x = 2 \pi n$, and $x = \pi + 2 \pi n$

$\sqrt{2} \sin x + 1 = 0$, $\implies$, $\sin x = - \frac{1}{\sqrt{2}}$, $\implies$, $x = \frac{5}{4} \pi + 2 \pi n$, and $x = \frac{7}{4} \pi + 2 \pi n$

$\sqrt{2} \sin x - 1 = 0$, $\implies$, $\sin x = \frac{1}{\sqrt{2}}$, $\implies$, $x = \frac{1}{4} \pi + 2 \pi n$ and $x = \frac{3}{4} \pi + 2 \pi n$

$\sin x + 1 = 0$, $\implies$, $\sin x = - 1$, $\implies$, $x = \frac{3}{2} \pi + 2 \pi n$

$\sin x - 1 = 0$, $\implies$, $\sin x = 1$, $\implies$, $x = \frac{1}{2} \pi + 2 \pi n$

Jul 16, 2017

0; pi/4; pi/2; pi

Explanation:

$f \left(x\right) = {\sin}^{2} 3 x - {\sin}^{2} x = \left(\sin 3 x - \sin x\right) \left(\sin 3 x + \sin x\right)$
Apply the 2 trig identities:
$\sin a - \sin b = 2 \cos \left(\frac{a + b}{2}\right) \sin \left(\frac{a - b}{2}\right)$
$\sin a + \sin b = 2 \sin \left(\frac{a + b}{2}\right) \cos \left(\frac{a - b}{2}\right)$
In this case -->
f(x) = (2cos 2x.sin x)(2sin 2x.cos x) = 0
f(x) = (2sin 2x.cos 2x)(2sin x.cos x) = sin 4x.sin 2x = 0
Either factor must be zero.

a. sin 2x = 0 --> 2x = 0; $2 x = \pi$, and $2 x = 2 \pi$ -->
x = 0; x = pi/2; and $x = \pi$
b. sin 4x = 0 --> 4x = 0; $4 x = \pi$; and $4 x = 2 \pi$-->
x = 0; x = pi/4; and $x = \frac{\pi}{2}$
Answers in interval $\left(0 , 2 \pi\right)$:
0; pi/4; pi/2; pi
Check by calculator:
$x = \frac{\pi}{4}$ --> $\sin x = \frac{1}{\sqrt{2}}$ --> ${\sin}^{2} x = \frac{1}{2}$ -->
$\sin 3 x = \sin \left(\frac{3 \pi}{4}\right) = \frac{\sqrt{2}}{2}$ --> ${\sin}^{2} \left(\frac{3 \pi}{4}\right) = \frac{2}{4} = \frac{1}{2}$.
${\sin}^{2} 3 x - {\sin}^{2} x = \frac{1}{2} - \frac{1}{2} = 0$ Proved.