Question

Stuck: trying to solve for x:

cos(2x)=-sin(x) where 0<=x<=2pi

My work below.

Answer 1

\[\cos^{2}x-\sin^{2}x=-\sin(x)\]

Answer 2

After that, a bunch of reshuffling that got me nowhere.

Answer 3

If you use \[\cos 2x = 1 - \sin ^{2} x\] you will have a quadratic in sin x

Answer 4

sorry, 1 - 2(sin x)^2

Answer 5

So, you use \[\cos 2x = 1 - 2\sin ^{2}x\]

Answer 6

cos(2x) is equal to 1 - 2sin^2(x)

This makes it all 1-2sin^2(x) = -sin(x)
re-arrange to make 2sin^2(x) - sin(x) - 1 =0

Answer 7

(2sinx-1)(sinx+1)

sinx = -1 or
sinx= 0.5

Answer 8

Would it have been easier if I had given the original problem? \[\cos(2x)=-\sin(x)\]

Answer 9

that's the same problem as your initial question, i was able to see what you mean

Answer 10

Oh geez, ignore. SO tired. Four hours sleep. LOL!

Answer 11

:D i know the feeling