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

Solution 1:

Hint: $ \sin^5 x\leq \sin^2 x$ and $ \cos ^3 x \leq \cos^2 x $.

Hint: Pythagorean Identity for trigonometric functions.

Solution 2:

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 $x = 2m\pi$.

Solution 3:

$\sin^5x\le \sin^2x, \cos^3x\le \cos^2x$, 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.