Math Help - Trig equation

1. Trig equation

I've found that in the first part of the question

$sin^5\theta = \frac{1}{16}(sin5\theta-5sin3\theta+10sin\theta)$

Now I have to solve $sin5\theta-5sin3\theta+6sin\theta = 0$ for theta

I can see straight away that 0 and multiples of pi are my solutions,
But I don't know how to do this problem in a methodical way.
I thought there would be substitution involved but I can't see how.

Help appreciated.

2. Re: Trig equation

You can do it in a long way (I replace $\theta$ by $x$).
To solve:
$\sin(5x)-5\sin(3x)+6\sin(x)=0$

First we're going to expand $\sin(5x)$ and $\sin(3x)$ in terms of $\sin(x)$, therefore:
$\sin(5x)=\sin(2x+3x)=\sin(2x)\cdot \cos(3x)+\cos(2x)\cdot \sin(3x)=...=16\sin^5(x)-20\sin^3(x)+5\sin(x)$
$\sin(3x)=\sin(x+2x)=...=3\sin(x)-4\sin^3(x)$

Therefore the equation becomes:
$16\sin^5(x)-20\sin^3(x)+5\sin(x)-15\sin(x)+20\sin^3(x)+6\sin(x)=0$
$\Leftrightarrow 16\sin^5(x)-4\sin(x)=0$
$\Leftrightarrow 4\sin(x)[4\sin^4(x)-1]=0$
$\Leftrightarrow ...$

3. Re: Trig equation

Which identities are you using?

4. Re: Trig equation

Expand $\sin(5x)$ in terms of $\sin(x)$, do the same for $\sin(3x)$.
Can you do that?
Hint: rewrite $\sin(5x)=\sin(2x+3x)$ and $\sin(3x)=\sin(2x+x)$ and use the formula: $\sin(a+b)=\sin(a)\cdot \cos(b)+\sin(b)\cdot \cos(a)$ and also where necessary $\cos^2(x)=1-\sin^2(x)$.

It will take some time to do that.

Thanks!

6. Re: Trig equation

You're welcome! Try to resolve the equation by your own calculations and offcourse take your time for doing that because in this case you have to pay attention with signs etc ... .

7. Re: Trig equation

$\sin^5\theta = \frac{1}{16}(\sin5\theta-5\sin3\theta+10\sin\theta)$

$16\sin^5\theta = \sin5\theta-5\sin3\theta+10\sin\theta$
$= (\sin5\theta-5\sin3\theta+6\sin\theta)+4\sin\theta$

$= (0)+4\sin\theta$

Therefore, $16\sin^5\theta = 4\sin\theta$ is equivalent to $\sin5\theta-5\sin3\theta+6\sin\theta=0\,.$