Solving $\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$ for $\alpha$

Question

I have been struggling with this equation. Is it unsolvable?

$$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$

$\beta$ and $c$ are constant. How can I solve for the angle $\alpha$?

PS: As you might imagine, I'm still new to this, so please be easy on me. Thanks for your help.

Answer 1

Rewrite $$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$ as $$ (1-c)\cos\alpha\cos\beta + (1+c)\sin\alpha\sin\beta = 0$$ Taking $a = (1-c)\cos\beta ~$ and $~b = -(1+c)\sin\beta$, we get $$a\cos\alpha = b\sin\alpha$$ Now, the case $~\cos\alpha = 0~$ is not interesting. Assume $\cos \alpha \ne 0$ and divide by it to get $$b\tan \alpha = a$$ Hence, if either of $a$ and $b$ is $0$, then so is another. Assume, neither is zero and divide by $b$ to get $$\tan \alpha = \frac ab = \frac{c - 1}{c + 1} \cot \beta$$

Now, you can decide if it is solvable or not.

Answer 2

I think I've found a solution to this problem, solving for $\alpha$. Following VIVID's answer i arrived at: $$\cos(\alpha)\cdot\cos(\beta)(1-c)-\sin(\alpha)\cdot\sin(\beta)(1+c)=0$$ following the angle sum formula: $$\cos(x+y)=\cos(x)\cdot\cos(y)-\sin(x)\cdot\sin(y)$$ finally i can express $\alpha$ in terms of $\beta$ and c, which gives...

$$\alpha(\beta,c) = \arctan\Biggr(\frac{1-c}{1+c}\cdot\frac{1}{\tan(\beta)}\Biggr)$$

I hope there is no big oversight and this is correct. Also, I've never used LaTeX before, so writing this was painfully slow, so I skipped some steps. I thank everyone for their help.