# Trigonometric equations

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

$\displaystyle \text{1.})$$\displaystyle sin x + sin 3x = 0 \displaystyle sin x + 3sin x - 4 sin^3 x = 0 \displaystyle -4sin x(sin^2x - 1) = 0 \displaystyle sin x = 0 \displaystyle x = {\pi}n \displaystyle sin^2x - 1 = 0 \displaystyle 1sin x = 1 \displaystyle x = {\pi} + 2 {\pi}k \displaystyle 2sinx = -1 \displaystyle x = - \frac{\pi}{2} + 2{\pi}k \displaystyle x = - \frac{\pi}{2} + 2{\pi}m \displaystyle x(1) = 0 \displaystyle x(2) = {\pi} \displaystyle x(3) = \frac{\pi}{2} \displaystyle 0\le x \le{\pi} \displaystyle \text{2.})sin{\theta} + sin2{\theta} + sin3{\theta} = 0 • Feb 9th 2010, 02:46 AM Prove It Quote: Originally Posted by Hellbent Am I correct on the first one? A little help with second one please. \displaystyle \text{1.})$$\displaystyle sin x + sin 3x = 0$

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

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

$\displaystyle sin x = 0$

$\displaystyle x = {\pi}n$

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

$\displaystyle \sin^2{x} = 1$

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

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

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

So putting it together you have

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

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

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

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

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

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

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

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

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

$\displaystyle \theta = \pi n$.

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

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

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

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

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

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

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

So putting it all together:

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