double angle question

**stasis** · May 20th 2008, 01:41 PM

How do you solve:

sin(2θ) + sin(4θ) = 0

on the interval of [0 degrees, 360 degrees)

**Moo** · May 20th 2008, 01:46 PM

Hello,

Originally Posted by stasis

How do you solve:

sin(2θ) + sin(4θ) = 0

on the interval of [0 degrees, 360 degrees)

Write : $\sin 4 \theta=2 \sin 2\theta \cdot \cos 2\theta$

Notice that $\cos A=\sqrt{1-\sin^2 A} \implies \cos 2\theta=\sqrt{1-\sin^2 2 \theta}$

Substitute : $y=\sin 2 \theta$

$\implies \sin 2 \theta+2 \sin 2 \theta \sqrt{1-\sin^2 2 \theta}=y+2y \cdot \sqrt{1-y^2}$

So you want to solve for y in $y+2y \cdot \sqrt{1-y^2}=0 \longleftrightarrow y(1+2\sqrt{1-y^2})=0$

Does it help ?

**Chris L T521** · May 23rd 2008, 05:26 PM

Originally Posted by stasis

How do you solve:

sin(2θ) + sin(4θ) = 0

on the interval of [0 degrees, 360 degrees)

Note that $\sin(2u)=2\sin(u) \cos(u)$ . Thus, $\sin(4\theta)=2\sin(2\theta) \cos(2\theta)$ . Substituting this into the equation, we get:

$\sin(2\theta)+2\sin(2\theta)\cos(2\theta)=0\implie s \sin(2\theta)\left(1+2\cos(2\theta)\right)=0$

Thus we get two equations:

$\sin(2\theta)=0 \ \text{and} \ \cos(2\theta)=-\frac{1}{2}$ . You can take it from here...

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