Question

How do you solve cot^2x + cscx= 1
(0,2pi)

Answer 1

`\bf 1+cot^2(\theta)=csc^2(\theta)\implies cot^2(\theta)=csc^2(\theta)-1 \\\\\\ csc(\theta)=\cfrac{1}{sin(\theta)} \\\\ -------------------------------\\\\ cot^2(x)+csc(x)=1\implies csc^2(x)-1+csc(x)=1 \\\\\\ csc^2(x)+csc(x)-2=0\implies [csc(x)-2][csc(x)+1]=0\\\\ -------------------------------\\\\`


`\bf csc(x)-2=0\implies csc(x)=2\implies \cfrac{1}{sin(x)}=2\implies \cfrac{1}{2}=sin(x) \\\\\\ sin^(-1)\left( (1)/(2) \right)=\measuredangle x\implies (\pi )/(6)~~,~~(5\pi )/(6)=\measuredangle x\\\\ -------------------------------\\\\ csc(x)+1=0\implies csc(x)=-1\implies \cfrac{1}{sin(x)}=-1 \\\\\\ \cfrac{1}{-1}=sin(x) \implies -1=sin(x)\implies sin^(-1)(-1)=\measuredangle x\implies (3\pi )/(2)=\measuredangle x`