# Find all the values of $\theta$ that satisfy $\cos(x \theta ) + \cos( (x+2) \theta ) = \cos( \theta )$

Find all the values of $$\theta$$ that satisfy the equation $$\cos(x \theta ) + \cos( (x+2) \theta ) = \cos( \theta )$$

I've tried simplifying with factor formulae and a combo of compound angle formulae, and I'm still stuck. I get to $$\theta = 180^\circ$$ and $$\theta = \frac{60^\circ}{x+1}$$, but I'm unsure if that's correct.

It seems to work for $$\theta=180^\circ$$, but I can't verify the other solution. I feel as though it should be a numerical solution, but I'm unsure.

• Is $x$ supposed to be an integer or any number? – Warren Hill Feb 20 at 13:18
• Great question. The question on the national examination is exactly as I posted it. Granted, the questions on the national examination in this country are riddled with mistakes, hence my reservations about my answer. – Suhly Feb 21 at 11:29

it's $$2\cos(x+1)\theta\cos\theta=\cos\theta$$ or $$(2\cos(x+1)\theta-1)\cos\theta=0.$$ Can you end it now?
Actually, $$\cos\theta=0$$ gives $$\theta=\frac{\pi}{2}+\pi k,$$ where $$k\in\mathbb Z$$.
I used $$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$$ and $$\pi=180^{\circ}.$$
Use $$\cos(p)+\cos(q)=2 \cos \left(\frac{p+q}{2}\right)\cos \left(\frac{p-q}{2}\right)$$ Make $$p=(x+2)\theta$$ and $$q=x\theta$$ making your problem to be $$2\cos((x+1)\theta) \cos(\theta)= \cos(\theta)$$ and then the two cases.