Question

Find all solutions of the equation in the interval [0,2pi).
2(sin^2)x = 2 + cosx

Answer 1

replace sin^2x with 1-cos^2x

Answer 2

Alright, I'm working it out in my notebook hold on

Answer 3

Do you get 2(cos^2)x -cosx = 0?

Answer 4

\[2(1-\cos ^2x)=2+\cos x\]
\[2-2\cos ^2x=2+\cos x\]
\[-2\cos ^2x-\cos x=0\]
\[\cos ^2x+\cos x=0\]

Answer 5

Whoops. Forgot the 2

Answer 6

\[2\cos ^2x+\cos x=0\]

Answer 7

Factor out cos x. Set each factor equal to 0 and solve.

Answer 8

Ah, thank you