Question

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

Answer 1

$\bf cos^4(x)=cos^4(x)csc(x)\\\\ -----------------------------\\\\ cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)} \\\\\\ cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0 \\\\\\ cos^4(x)[sin(x)-1]=0\to \begin{cases} cos^4(x)=0\to x=cos^{-1}(0)\\ ----------\\ sin(x)-1=0\\ sin(x)=1\to x=sin^{-1}(1) \end{cases}$

$\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases} \\\\\\ \measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}$