Question

Solve the following question:
(-sin^2)x=2cosx-2

Answer 1

Start by moving everything to one side

Answer 2

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

Answer 3

okay so would that be the final answer?
im kinda confused I dont know if should be proofing it or solving for x

Answer 4

no, the final answer is: \[cosx=\pm1\]
where cos equals +/-1

Answer 5

so what trig values equal +/-1?

Answer 6

would you do x=(cos^-1)(1)

Answer 7

you would find the inverse

Answer 8

You could.. you could also use the unit circle and see where cos is 1/-1