General solution of the equation $\sin^{2015}(x)+\cos^{2015}(x) = 1\; \forall \; x\in \mathbb{R}$ 
Calculation of General solution of the equation $\sin^{2015}(x)+\cos^{2015}(x) = 1\; \forall \;x\in \mathbb{R}$.

$\bf{My\; Try:}$ We can Write the equation as $$\sin^{2015}(x)+\cos^{2015}(x)\leq \sin^2(x)+\cos^2(x)=1.$$
And equality hold when $$\sin(x)=1$$ and $$\cos (x) = 0.$$
So we get $$\displaystyle x=n\pi+(-1)^n\cdot \frac{\pi}{2}$$ and $$\displaystyle x=2n\pi\pm \frac{\pi}{2}.$$
Now How can I calculate common solution of $\sin (x) = 0$ and $\cos (x) = 0$.
Please help me. Thanks.
 A: Your approach seems correct although $\sin(x)=\cos(x)=0$ is just nonsense. You have $(a,b)=(\cos x,\sin x)\in[-1,1]^2$ satisfying
$$
a^{2015}+b^{2015}\leq |a|^{2015}+|b|^{2015}\leq a^2+b^2=1
$$
But if $|a|,|b|\in(0,1)$ we have $|a|^{2015}<a^2$ and $|b|^{2015}<b^2$ so in that case the inequality above must be strict. So only for $\{a,b\}=\{0,1\}$ this can hold.
Correction: This means either $a=\cos(x)=1$ or $b=\sin(x)=1$ which will give you all solutions. Before I said $\cos(x)=0$ would work, but then $\sin(x)=\pm 1$ and not $\sin(x)=1$ as desired.
A: $\sin^{2015}(x) + \cos^{2015}(x) = 1 = \sin^2(x) + \cos^2(x)$
or, $\left[\sin^2(x) \left(\sin^{2013}(x)-1 \right) \right]+ \left[ \cos^2(x) \left( \cos^{2013}(x)-1 \right) \right]= 0$
Since $\sin^2(x) \geq 0$ and $\cos^2(x) \geq 0$, this is possible only if both
$\left[\sin^2(x) \left(\sin^{2013}(x)-1 \right) \right]= 0$
and 
$\left[ \cos^2(x) \left( \cos^{2013}(x)-1 \right) \right]= 0$
That is possible if either


*

*$\sin^2(x) = 0$ and $\left( \cos^{2013}(x)-1 \right) = 0$
$\implies \sin(x) = 0$ and $\cos(x) = 1 \implies x = 2n \pi$
or


*$\cos^2(x) = 0$ and $\left(\sin^{2013}(x)-1 \right) = 0$
$\implies \cos(x) = 0$ and $\sin(x) = 1 \implies x = (4n+1)\frac{\pi}{2}$
