Stuck: trying to solve for x:

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

Question

Stuck:  trying to solve for x:

cos(2x)=-sin(x) where 0<=x<=2pi

My work below.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

sorry, 1 - 2(sin x)^2

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

(2sinx-1)(sinx+1)

sinx = -1  or
sinx= 0.5

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

:D i know the feeling