How many solutions are there of the equation $(\cos a)^x+(\sin a)^x=1$, $x \in \mathbb{R}$. Is there any solution of the equation other than $x=2$?
Please help me. Thank you in advance.
 A: First important thing to consider is that, when $\cos a$ or $\sin a$ are negative the exponentiation should be done with care, because it is in general multi-valued. For instance, if $\cos a=-1$, then :
$$
(-1)^x=e^{i(2k+1)\pi\, x}.
$$
In this case, we have a multi-valued expression giving different values for all $k\in\mathbb Z$. However we can always take a general $m$ and $n$, not necessarily equal, for which we have:
$$
\cos a=|\cos a|e^{im\pi}\,\,\,\,\text{   and   }\,\,\,\, \sin a=|\sin a|e^{in\pi}.
$$
So the equation is equivalent to finding $x$ such that for some $m,n\in\mathbb Z$, we have:
$$
|\cos a|^xe^{imx\pi}+|\sin a|^xe^{inx\pi}=1.
$$
Taking the norm from both side and using triangle inequality we can see that :
$$
|\cos a|^x+|\sin a|^x\geq 1.
$$
If $|\cos a|$ and $\sin a$ are both strictly positive, the function is strictly increasing and therefore the answer is only $x=2$. However if one of them is zero and hence the other one is one, the inequality also holds. So we have to check the cases where $|\cos a |=1$ or $|\sin a|=1$. 


*

*When $\cos a=1$ or $\sin a =1$, all $x\in\mathbb R$ is an answer.

*When $\cos a=-1$ or $\sin a=-1$, then we have:
$$
(-1)^x=1\implies e^{(2p+1)\pi x i}=1\implies (2p+1)\pi x=2q\pi,
$$
so all $x=\frac{2q}{2p+1}$ are answers for $p,q\in\mathbb Z$.
