How can I find all the solutions of $\sin^5x+\cos^3x=1$ 
Find all the solutions of $$\sin^5x+\cos^3x=1$$

Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.
 A: Use what you know about the magnitudes of $\sin x$ and $\cos x$.
$$
\begin{aligned}
\sin^5x+\cos^3x &\le |\sin^5x+\cos^3x| \\
&\le |\sin^5 x| + |\cos^3 x| \\
&\le |\sin^2 x| + |\cos^2 x| \\
&= \sin^2 x + \cos^2 x\\
&= 1
\end{aligned}
$$
The inequality where the exponents are changed is only satisfied if the individual terms are equal: $\sin x$ and $\cos x$ must both be $0$ or $1$. Put that into the original equation, and you get $\sin x = 1$ or $\cos x = 1$. So,  $x= \frac\pi2 + 2n\pi$ or $x = 2m\pi$.
A: Hint: $ \sin^5 x\leq \sin^2 x$ and $ \cos ^3 x \leq \cos^2 x $.
Hint: Pythagorean Identity for trigonometric functions.
A: $\sin^5x\le \sin^2x, \cos^3x\le \cos^2x$, and $\sin^2x+\cos^2x=1$, so $\sin^5x+\cos^3x$
can be equal to 1 if and only if $$\sin^5x=\sin^2x$$ and $$\cos^3x=\cos^2x$$, i.e.
$$\sin^2x(1-\sin^3x)=0$$ and $$\cos^2x(1-\cos x)=0$$
The first equation has solution: $x=k\pi$ or $x=\pi/2+2m\pi, k,m $ integers.
The second equation has solution: $x=2n\pi$ or $x=\pi/2+p\pi, n,p$ integers.
Therefore the solution is either:
(i) $x=k\pi$ and $x=2n\pi$, or equivalently, $x=2n\pi$ for some integer $n$, or 
(ii) $x=k\pi$ and $x=\pi/2+p\pi$ (impossible), or
(iii) $x=\pi/2+2m\pi$ and $x=2n\pi$ (impossible) or
(iv) $x=\pi/2+2m\pi$ and $x=\pi/2+p\pi$, or equivalently, $x=\pi/2+2m\pi$ for some integer $p$.
In conclusion, $x=2n\pi$ or $x=\pi/2+2n\pi$, $n$ integers, are the solutions.
A: By letting $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}$ with $t=\tan(x/2)$, we find that the given equation is equivalent to
$$t^2(t-1)^2\underbrace{((t^2+1)(t^2+3)(t+1)^2+8t^2)}_{>0}=0$$
By solving $t=0$ and $t=1$, it follows easily that the solutions are
$$x_n=2n\pi\quad,\quad x'_n=\pi/2+2n\pi \quad \text{with $n\in\mathbb{Z}$}.$$
