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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.

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  • $\begingroup$ Is $x$ supposed to be an integer or any number? $\endgroup$ – Warren Hill Feb 20 at 13:18
  • $\begingroup$ 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. $\endgroup$ – Suhly Feb 21 at 11:29
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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 $k\in\mathbb Z$.

Also, there is a mistake in your second sequence.

I used $$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$$ and $$\pi=180^{\circ}.$$

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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.

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