$2\sin(x)=\sin(3x)$ .Solve for $x$.

Question

I got this equation as a part of the sum that I am solving.Though I know the ans for $\sin(x)=\sin(a)$. But don't know how to get this one as there is a 2 in multiplication

Answer 1

Hint:

Solve $$2\sin { \left( x \right) } =\sin { \left( 3x \right) } \\ 2\sin { \left( x \right) =3\sin { \left( x \right) -4\sin ^{ 3 }{ \left( x \right) } } } \\ \sin { \left( x \right) \left( 4\sin ^{ 2 }{ \left( x \right) -1 } \right) =0 } $$

Answer 2

Well, we know that:

• $$\sin\left(3x\right)=3\sin\left(x\right)\cos^2\left(x\right)-\sin^3\left(x\right)\tag1$$
• $$\cos^2\left(x\right)+\sin^2\left(x\right)=1\tag2$$

So, we get:

$$2\sin\left(x\right)=\sin\left(3x\right)=3\sin\left(x\right)\cdot\left(1-\sin^2\left(x\right)\right)-\sin^3\left(x\right)\tag3$$

Let $\text{u}=\sin\left(x\right)$:

$$2\text{u}=3\text{u}\cdot\left(1-\text{u}^2\right)-\text{u}^3\space\Longleftrightarrow\space\text{u}=0\space\vee\space\text{u}=\pm\space\frac{1}{2}\tag4$$

Answer 3

Hint:

$$\sin x=\sin3x-\sin x=2\sin x\cos2x$$

$$\iff\sin x(2\cos2x-1)=0$$