Question

solve sin^2x-5cosx=5 interval [0,2pi)

Answer 1

\(\large\color{black}{ \sin^2x-5\cos x=5 }\)
\(\large\color{black}{ (1-\cos^2x)-5\cos x=5 }\)
\(\large\color{black}{ 1-\cos^2x-5\cos x=5 }\)
\(\large\color{black}{ -\cos^2x-5\cos x=6 }\)
\(\large\color{black}{ \cos^2x+5\cos x=-6 }\)
\(\large\color{black}{ \cos^2x+5\cos x+6=0 }\)
\(\large\color{black}{ (\cos x+2)(\cos x+3)=0 }\)

Answer 2

\[\sin^2x+\cos^2x=1\]

so you can replace sin^2x with (1-cos^x)

Answer 3

\[\large\color{black}{ (1-\cos^2x)-5\cos x=5 }\]\[\large\color{black}{ 1-\cos^2x-5\cos x=5 }\]\[\large -\cos^2x-5\cos x=\color{red}4 \]

Answer 4

\[\large \cos^2x+5\cos x+4 =0 \]\[\large (\cos x+1)(\cos x+4) =0 \]

so, \(\cos x=-4\) which is not possible.
.....or \(\cos x=-1\) which happens at \(x=\color{green}\pi\)