General sloution but requires manipulation

**dojo** · November 5th 2010, 12:07 PM

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

**Unknown008** · November 5th 2010, 12:18 PM

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?

**Soroban** · November 5th 2010, 01:11 PM

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

Math Help - General sloution but requires manipulation

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