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} (\pi )/(2)\\\\ (3\pi )/(2) \end{cases} \\\\\\ \measuredangle x=sin^(-1)(1)\implies \measuredangle x=(\pi )/(2)`