# Trigonometric equations

• February 9th 2010, 02:14 AM
Hellbent
Trigonometric equations
Am I correct on the first one?
A little help with second one please.

$\text{1.})$ $sin x + sin 3x = 0$

$sin x + 3sin x - 4 sin^3 x = 0$

$-4sin x(sin^2x - 1) = 0$

$sin x = 0$

$x = {\pi}n$

$sin^2x - 1 = 0$

$1sin x = 1$

$x = {\pi} + 2 {\pi}k$

$2sinx = -1$

$x = - \frac{\pi}{2} + 2{\pi}k$

$x = - \frac{\pi}{2} + 2{\pi}m$

$x(1) = 0$

$x(2) = {\pi}$

$x(3) = \frac{\pi}{2}$

$0\le x \le{\pi}$

$\text{2.})sin{\theta} + sin2{\theta} + sin3{\theta} = 0$
• February 9th 2010, 02:46 AM
Prove It
Quote:

Originally Posted by Hellbent
Am I correct on the first one?
A little help with second one please.

$\text{1.})$ $sin x + sin 3x = 0$

$sin x + 3sin x - 4 sin^3 x = 0$

$-4sin x(sin^2x - 1) = 0$

$sin x = 0$

$x = {\pi}n$

$sin^2x - 1 = 0$ Correct up to here...

$\sin^2{x} = 1$

$\sin{x} = \pm 1$...

So $x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n$

$x = \frac{\pi}{2} + \pi n$.

So putting it together you have

$x = \left\{0, \frac{\pi}{2}\right\} + \pi n$.
• February 9th 2010, 02:56 AM
Prove It
Quote:

Originally Posted by Hellbent
$\text{2.})sin{\theta} + sin2{\theta} + sin3{\theta} = 0$

$\sin{\theta} + 2\sin{\theta}\cos{\theta} + 3\sin{\theta} - 4\sin^3{\theta} = 0$

$4\sin{\theta} + 2\sin{\theta}\cos{\theta} - 4\sin^3{\theta} = 0$

$\sin{\theta}(4 + 2\cos{\theta} - 4\sin^2{\theta}) = 0$

$\sin{\theta}[4(1 - \sin^2{\theta}) + 2\cos{\theta}] = 0$

$\sin{\theta}(4\cos^2{\theta} + 2\cos{\theta}) = 0$

$2\sin{\theta}\cos{\theta}(2\cos{\theta} + 1) = 0$

Case 1: $\sin{\theta} = 0$

$\theta = \pi n$.

Case 2: $\cos{\theta} = 0$

$\theta = \frac{\pi}{2} + \pi n$

Case 3: $2\cos{\theta} + 1 = 0$

$2\cos{\theta} = -1$

$\cos{\theta} = -\frac{1}{2}$

$\theta = \left\{ \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3} \right\} + 2\pi n$

$\theta = \left \{ \frac{2\pi}{3}, \frac{4\pi}{3} \right\} + 2\pi n$.

So putting it all together:

$\theta = \left \{ 0, \frac{\pi}{2}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{3\pi}{2} \right\} + 2\pi n$.
• February 9th 2010, 03:33 AM
Hellbent
Ah! Thanks for your help :)
• February 9th 2010, 04:54 AM
Krizalid
having $2\sin\theta\cos\theta(2\cos\theta+1)=0,$ we have that this equals $\sin(2\theta)(2\cos\theta+1)=0$ so $\theta=\frac{k\pi}2$ and the other solutions already stated by Prove It.