# Math Help - General sloution but requires manipulation

1. ## General sloution but requires manipulation

I'm have troubles with finding the general solution for [tex] \theta [\math] of:
$\cos(6\theta) + \cos(3\theta) + 4\cos^2 (\frac{3}{2}\theta)sin^2(\frac{3}{2}\theta) = 0$

I know the cosine general formula is this:
$\pm\theta +2n\pi$
and that you can use identities to break the functions down until each factor can be zeroed and hence find these values. But the two additions are throwing me at the start amongst other things.

Your help would be greatly appreciated.

D

2. You could start by using the double angle identity for sine.

$\sin2A = 2\sin A\cos A$

Squaring both sides give:

$\sin^22A = 4\sin^2A\cos^2A$

Subtituting A by 3theta/2;

$\sin^22(\dfrac{3\theta}{2}) = 4\sin^2(\dfrac{3\theta}{2})\cos^2(\dfrac{3\theta}{ 2})$

$\sin^23\theta = 4\sin^2(\dfrac{3\theta}{2})\cos^2(\dfrac{3\theta}{ 2})$

Now we see that we have cos 3theta and cos 6theta. Break the cos 6theta into half angle.

$\cos2A = 2\cos^2A - 1$

$\cos6\theta = 2\cos^23\theta - 1$

Can you finish this now?

3. Hello, dojo!

$\cos 6\theta + \cos3\theta + 4\cos^2(\tfrac{3}{2}\theta)\sin^2(\tfrac{3}{2}\the ta) \:=\: 0$

$\text{We have: }\;\cos6\theta + \cos3\theta + \underbrace{\bigg[2\sin(\tfrac{3}{2}\theta)\cos(\tfrac{3}{2}\theta)\ bigg]^2} \;=\;0$
. . . . . $\overbrace{(2\cos^2\!3\theta - 1)} \;+\; \cos3\theta \;+\; \sin^23\theta \;=\;0$

. . . . . $2\cos^23\theta - 1 + \cos3\theta + \overbrace{(1 - \cos^23\theta)} \;=\;0$

. . . . . . . . . . . . . . . . . . . . $\cos^23\theta + \cos3\theta \;=\;0$

. . . . . . . . . . . . . . . . . . $\cos3\theta(\cos3\theta +1) \;=\;0$

Therefore: . $\begin{Bmatrix}\cos3\theta \:=\: 0 & \Rightarrow & 3\theta \:=\: \frac{\pi}{2} + \pi n & \Rightarrow & \theta \:=\: \frac{\pi}{6} + \frac{\pi}{3}n \\ \\[-3mm]
\cos3\theta\:=\:\text{-}1 & \Rightarrow & 3\theta \:=\:\pi + 2\pi n & \Rightarrow & \theta \:=\:\frac{\pi}{3} + \frac{2\pi}{3}n

\end{Bmatrix}$