Question

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

Answer 1

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

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