# A trigonometric equation $\left(\sin{\frac{\pi}{7}}\right)^x+\left(\cos{\frac{\pi}{7}}\right)^x=1$

What's the way to solve $$\left(\sin{\frac{\pi}{7}}\right)^x+\left(\cos{\frac{\pi}{7}}\right)^x=1$$ I am looking for an analytic solution for this equation. With numerical solving I can find the solution(s), but if possible guide me to solve it like a real man!
remark: I know $x=2$ works here .

Notice that $0 < \sin{\frac{\pi}{7}} , \ \cos{\frac{\pi}{7}} < 1$;

note that for every $2 < x$ we can conclude that:

$$(\sin{\frac{\pi}{7}})^x+(\cos{\frac{\pi}{7}})^x < (\sin{\frac{\pi}{7}})^2+(\cos{\frac{\pi}{7}})^2 =1 \Longrightarrow \\ (\sin{\frac{\pi}{7}})^x+(\cos{\frac{\pi}{7}})^x < 1 \ \ \ \ \ \ \ \ \text{for every} \ \ \ \ 2 < x \in \mathbb{R} ;$$

also for every $x < 2$ we can conclude that:

$$(\sin{\frac{\pi}{7}})^x+(\cos{\frac{\pi}{7}})^x > (\sin{\frac{\pi}{7}})^2+(\cos{\frac{\pi}{7}})^2 =1 \Longrightarrow \\ (\sin{\frac{\pi}{7}})^x+(\cos{\frac{\pi}{7}})^x > 1 \ \ \ \ \ \ \ \ \text{for every} \ \ \ \ 2 > x \in \mathbb{R} ;$$

so there is no other solution rather than $x=2$.

It has one solution: $x=2$.

Besides, $\left(\sin{\frac\pi7}\right)^x$ and $\left(\cos{\frac\pi7}\right)^x$ are strictly decreasing functions and therefore so is their sum. So, $2$ is the only solution.

The only solution is $x=2$ which comes from the fundamental identity $\sin^2\alpha+\cos^2\alpha=1,\;\forall\alpha\in\mathbb{R}$

To show that there are no other solutions I computed the first derivative of

$f(x)=\sin ^x\left(\frac{\pi }{7}\right)+\cos ^x\left(\frac{\pi }{7}\right)-1$

which is

$f'(x)=\sin ^x\left(\frac{\pi }{7}\right) \log \left(\sin \left(\frac{\pi }{7}\right)\right)+\cos ^x\left(\frac{\pi }{7}\right) \log \left(\cos \left(\frac{\pi }{7}\right)\right)$

as both log are negative because sine and cosine are less than $1$ we can conclude that $f(x)$ is decreasing on $\mathbb{R}$ therefore $x=2$ is the only solution

Hope this helps