# Trigonometric equations - Find the general solution

1. Jul 14, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
$$\cos x-\sqrt{3}\sin x=\cos(3x)$$

2. Relevant equations

3. The attempt at a solution
Dividing both the sides by two i.e
$$\cos x \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3}=\cos (3x)/2$$
LHS can be written as $\cos(x+\pi/3)$. Substituting $x+\pi/3=t \Rightarrow 3x=3t-\pi \Rightarrow \cos(3x)=-\cos(3t)$
$$\cos t=-\cos(3t)/2 \Rightarrow 2\cos t+\cos(3t)=0$$
$\because \cos(3t)=4\cos^3t-3\cos t$
$$4\cos^3t-\cos t=0 \Rightarrow \cos t(4\cos^2t-1)=0$$
Hence, $\cos t=0 \Rightarrow t=(2n+1)\pi/2 \Rightarrow x=(2n+1)\pi/2-\pi/3$ and $\cos t=±1/2 \Rightarrow t=k\pi ± \pi/3 \Rightarrow x=k\pi, k\pi-2\pi/3$ where $n,k \in Z$.

But these are wrong. The correct answers are: $n\pi, \pi(3k+(-1)^k)/6$.

Any help is appreciated. Thanks!

2. Jul 14, 2013

### ehild

It is too complicated. Try to expand cos(3x)=cos(x+2x) in the original equation.

ehild

3. Jul 14, 2013

### Pranav-Arora

I already devised a way to reach the correct answer. What I have posted above is my first attempt at the problem and I am curious as to where I went wrong with that.

To reach the correct answer, I would rewrite the given equation as $\cos(3x)-\cos x+\sqrt{3}\sin x=0$. Rewriting $\cos(3x)-\cos x$ as $-2\sin(2x)\sin x$, I can easily reach the correct answer.

4. Jul 14, 2013

### LCKurtz

Where did that equation I have highlighted in red come from?

Last edited: Jul 14, 2013
5. Jul 14, 2013

### verty

Pranav, I followed exactly your method and got the right answer. Your mistake must be after the word "hence". I got (using degrees) x = 180k + (0,30,60).

6. Jul 14, 2013

### haruspex

Yes, both answers reduce to $n\pi, n\pi+\pi/6, n\pi+\pi/3$