Hi Guys,

I am trying to investigate a function, during the process I need to solve this equation:

sin(x)+cos(2x)=0

in the region 0<=x<=2*Pai

How do I solve it ?

thanks !

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- Mar 5th 2011, 06:23 AMWeeGhow to solve this equation ?
Hi Guys,

I am trying to investigate a function, during the process I need to solve this equation:

sin(x)+cos(2x)=0

in the region 0<=x<=2*Pai

How do I solve it ?

thanks ! - Mar 5th 2011, 06:25 AMProve It
Use the identity $\displaystyle \displaystyle \cos{2x} \equiv 1 - 2\sin^2{x}$ to turn your equation into a quadratic, which you can then solve.

- Mar 5th 2011, 06:34 AMTheCoffeeMachine
Start by writing it as $\displaystyle \cos{2x} = -\sin{x}$ and recall that $\displaystyle \cos\left(\varphi+\frac{\pi}{2}\right) = -\sin{\varphi}$.

So we have $\displaystyle \cos{2x} = \cos\left(x+\frac{\pi}{2}\right)$. You should be able to solve it with ease now.

**Hint:**if $\displaystyle \cos{\theta} = \cos{\alpha}$ then $\displaystyle \theta = 2n\pi\pm \alpha$ (where $\displaystyle n$ is an integer, or zero). - Mar 5th 2011, 07:11 AMWeeG
I used the quadratic idea, I did t=sin(x)

and got t1=-0.5 and t2=1

thus, I got:

sin(x)=-0.5 and sin(x)=1 my solution is x=pai/2 and x=11/6*pai

but there should be another one at x=3.63, I have no idea why.... - Mar 5th 2011, 07:31 AMskeeter