The number of solutions of ^7x+^7x=1, x[0,4] is equal to:

Question

The number of solutions of $\sin^7x+\cos^7x=1$, $x\in[0,4\pi]$ is equal to:

Accepted Answer

Since $\sin^7x\le\sin^2x$ and $\cos^7x\le\cos^2x$ whenever $\sin x,\cos x\in[-1,1]$ (with care for signs) and $\sin^2x+\cos^2x=1$, equality forces $\sin^7x=\sin^2x$ and $\cos^7x=\cos^2x$, which happens only when $\sin x\in\{0,1\}$ and correspondingly $\cos x\in\{1,0\}$.

So $\sin x=0,\cos x=1$ or $\sin x=1,\cos x=0$, giving $x = 0,2\pi,4\pi,\dfrac\pi2,\dfrac{5\pi}2$ --- 5 solutions.

Solutions · 1