Question

2cos^2 x-1=-sinx

Answer 1

are you asking to prove the identity
\[\Large 2 cos^2x - 1 = -sin x\]

are you sure that you type the question correctly?:)

Answer 2

Replace cos^2 with 1-sin^2

Answer 3

The question was stated "Find all solutions in [0, 2π ) and also the general solution:"

Answer 4

ahh.., I see..., I thought you're asking to prove it... :D

now.., first replace the \( cos^2 x\) as @Mertsj said...

what do you get?

Answer 5

I'm a little confused..?

Answer 6

\[2(1-\sin ^2x)-1=-\sin x\]

Answer 7

Now use the distributive property, get it equal to 0 and factor it.

Answer 8

\[\large 2\cos^2 x-1=-\sin x\]
Using your trig identities:
\[\sin^2 x+\cos^2 x=1\]
\[\cos^2 x=1-\sin^2 x\]
\[\large 2(1-\sin^2 x)-1=-\sin x\]
\[\large 2-2\sin^2 x-1=-\sin x\]
Move everything to the right hand side:
\[\large 2\sin^2 x-\sin x-1=0\]
\[\large (2\sin x+1 )(\sin x-1 )=0\]
\[\large \sin x=-\frac{1}{2}, 1\]
\[\large x=210, 330, 90\]

Answer 9

\[\large x=\frac{\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6}\]

Answer 10

@kausarsalley