Trigonometric equations - Find the general solution

[Saitama]

Homework Statement

$$\cos x-\sqrt{3}\sin x=\cos(3x)$$

Homework Equations

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!

 

[ehild]

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

ehild

 

[Saitama]

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

ehild

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.

 

[LCKurtz]

Homework Statement

$$\cos x-\sqrt{3}\sin x=\cos(3x)$$

Homework Equations

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 $x+\pi/3=t \Rightarrow 3x=3t-\pi \Rightarrow \cos(3x)=-\cos(3t)$
$$\cos t=-\cos(3t)/2 \Rightarrow 2\cos t+\cos(3t)=0$$
$\because \color{red}{\cos(3t)=4\cos^3t-3\cos t}$
$$4\cos^3t-\cos t=0 \Rightarrow \cos t(4\cos^2t-1)=0$$
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!

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

[Edit - added] Never mind about the identity. I think your work is correct and your answer and the "correct" answer are identical.

 

[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).

 

[haruspex]

I think your work is correct and your answer and the "correct" answer are identical.

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