Trigonometric Equation

**WolfMV** · December 5th 2008, 10:10 AM

Question please everyone, I'm a bit stumped on this problem. For it I have to find the general solutions.

The problem is:

2cos^2(x) = 1 + sin(x)

I'm just stumped as to how to solve the problem, do I use a power reducing formula or?

Thanks everyone!

**Sean12345** · December 5th 2008, 10:19 AM

Hi WolfMV,

Note that $\sin^2 x+\cos^2 x=1$ and so $\cos^2 x=1-\sin^2 x$ thus the equation $2\cos^2 x=1+\sin x$ becomes $2-2\sin^2 x=1+\sin x$ . Rearranging gives $2\sin^2 x+\sin x-1=0$ and we can see that $(2\sin x-1)(\sin x +1)=0$ .

Hope this helps.

**nzmathman** · December 5th 2008, 10:19 AM

Hint:
Use the trig identity $\cos{(2x)} = 1 - 2\sin^2{(x)}$
A quadratic equation will be formed which you can solve to find x.

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