how to solve this equation ?

**WeeG** · March 5th 2011, 06:23 AM

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 !

**Prove It** · March 5th 2011, 06:25 AM

Use the identity $\displaystyle \cos{2x} \equiv 1 - 2\sin^2{x}$ to turn your equation into a quadratic, which you can then solve.

**TheCoffeeMachine** · March 5th 2011, 06:34 AM

Start by writing it as $\cos{2x} = -\sin{x}$ and recall that $\cos\left(\varphi+\frac{\pi}{2}\right) = -\sin{\varphi}$ .
So we have $\cos{2x} = \cos\left(x+\frac{\pi}{2}\right)$ . You should be able to solve it with ease now.
Hint: if $\cos{\theta} = \cos{\alpha}$ then $\theta = 2n\pi\pm \alpha$ (where $n$ is an integer, or zero).

**WeeG** · March 5th 2011, 07:11 AM

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....

**skeeter** · March 5th 2011, 07:31 AM

Originally Posted by WeeG

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....

$\sin{x} = -\dfrac{1}{2}$ at $x = \dfrac{7\pi}{6}$ and $x = \dfrac{11\pi}{6}$

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