Question

how do you solve csc^2x+2cotx=5

Answer 1

2cotx = (csc^2x)(sin2x)

sin2x = 2sinxcosx

csc^2x = 1 / sin^2x

2cotx = 2sinxcosx / sin^2x

2cotx = 2cosx / sinx >>>2cotx

Answer 2

Thats it

Answer 3

Where did the 5 in the problem go?

Answer 4

I am not sure how to do it:/ Best of luck.

Answer 5

\[\csc ^{2}x+2\cot x-5=0\]
\[1+\cot ^{2}x+2\cot x-5=0,\cot ^{2}x+2\cot x-4=0\]
\[cotx =\frac{ -2\pm \sqrt{4-4*1*-4} }{2*1 }\]
\[\cot x=\frac{ -2\pm \sqrt{20} }{2 }=\frac{ -2\pm2\sqrt{5} }{2 }=-1\pm \sqrt{5}\]