Solving $\sin x = 4\sin10°\sin40°\sin(70°-x)$ So, I have this equation:
$$\sin x = 4\sin10°\sin40°\sin(70°-x)$$
And I'm trying to solve for $x$. Apparently $x=20°$ is the (smallest positive) solution but I can't arrive at it. I'm not very comfortable with this kind of equation so I just try to simplify it using prosthaphaere-something formulas, yet I don't get anywhere but to a messier equation to deal with.
Can someone solve it or at least point me in the right direction? Also, if you could tell me where I can find some material on solving equations like this (I can only find high school trig), that'd be great. Thanks.
 A: Note: All the work here, is in degrees.
First off all it is easy to prove this identity:
$\sin(3x)= 4\sin(x)\cdot \sin(60^{\circ}-x) \cdot \sin(60^{\circ}+x)$. 
Now, $$ \sin x = 4\sin10°\sin40°\sin(70°-x) $$
$$ \sin x = \frac{4\sin10°\sin50°\sin(70^{\circ})}{\sin(50)\sin(70)} \cdot \sin(40)\cdot \sin(70-x)= \frac{\sin(30)\sin(40)\sin(70-x)}{\sin(50)\sin(70)}$$
$$ \frac{\sin(30)\sin(40)\sin(70-x)}{\sin(50)\sin(70)} = \frac{\sin(30)\sin(40)\sin(70-x)}{\sin(50)\cos(20)} = \frac{\sin(20)\sin(70-x)}{\sin(50)} $$
So we are left with,  $\sin(x) \sin(50) = sin(20) \sin(70-x)$ . Its clear that $x=20$.One can prove it is only solution, by expanding ..
$$\sin(x)\sin(50) = \sin(20)\sin(70)\cos(x)-\sin(20)\sin(x)\cos(70)$$
$$ \sin(70) \cos(20) - \sin(20)\cos(70) = \sin(70-20)= \sin(20)\sin(70)\cot(x)-\sin(20)\sin(70) $$ 
Thus $ \tan(x)  = \tan(20) $
A: Using Werner Formulas,
$$2\sin10\sin40=\cos30-\cos50$$
Again, $$4\sin10\sin40\sin(70-x)=2(\cos30-\cos50)\sin(70-x)$$
$$=\sin(100-x)+\sin(40-x)-[\sin(120-x)+\sin(20-x)]$$
Now,
$$\sin x=4\sin10\sin40\sin(70-x)$$
$$\iff\sin x=\sin(100-x)+\sin(40-x)-[\sin(120-x)+\sin(20-x)] $$
$$\iff\sin x-\sin(40-x)=[\sin(100-x)-\sin(120-x)]+\sin(x-20) $$
$$\iff2\sin(x-20)\cos20=-2\sin10\cos(110-x)+\sin(x-20)$$
As $\displaystyle\cos(110-x)=\cos\{90-(x-20)\}=\sin(x-20),$ this becomes
$$\sin(x-20)(2\cos20-1)=-2\sin10\sin(x-20)$$
$$\iff\sin(x-20)(2\cos20-1+2\sin10)=0$$
But $\displaystyle\cos20+\sin10=\cos20+\cos80=2\cos50\cos30=\sqrt3\cos50\ne1$
$\displaystyle\implies\sin(x-20)=0\implies x-20=180^\circ n+20$ where $n$ is any integer
A: A way to address such trigonometric equations is to express everything in terms of the sine and cosine of the unknown. In your case, you can get rid of the argument $70°-x$ using the angle subtraction formula.
So, develop $$\sin x=4\sin10°\sin40°\sin70°\cos x-4\sin10°\sin40°\cos70°\sin x.$$
Doing that, you obtain a linear trigonometric equation of the form
$$A\cos x+B\sin x=C,$$ for which a general solution is known.
But your case is simpler, as $C=0$, and you directly derive 
$$\tan x=\frac{4\sin10°\sin40°\sin70°}{1+4\sin10°\sin40°\cos70°}.$$
Using a calculator, you find $x=20°$.
As $\sin x=0$ is not a solution, you didn't change the equation by this transformation. And given that the tangent function has a half turn period, the complete solution is
$$x=20°+180°k.$$
Now you easily prove that $x=20°$ is an exact solution with
$$4\sin10°(\sin40°\sin50°)=4\sin10°\frac12(\cos10°-\cos90°)=2\sin10°\cos10°=2\frac12(\sin20°-\sin0°)=\sin20°,$$
by two applications of the the prosthaphaeresis formulas.
A: Hint: Use the fact that $\sin{x-y}=\sin x \cos y-\cos x \sin y$ to expand $\sin(70°-x)$. From there you should be able to solve for $x$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sin\pars{x} = 4\sin\pars{10\c}\sin\pars{40\c}\sin\pars{70\c - x}:\
     {\large ?}}$

\begin{align}
&4\sin\pars{10\c}\sin\pars{40\c}
={\sin\pars{35\c}\cos\pars{35\c - x} - \cos\pars{35\c}\sin\pars{35\c - x}\over
\sin\pars{35\c}\cos\pars{35\c - x} + \cos\pars{35\c}\sin\pars{35\c - x}}
\\[3mm]&={1 - \cot\pars{35\c}\tan\pars{35\c - x}\over
1 + \cot\pars{35\c}\tan\pars{35\c - x}}
\\[3mm]&\imp\quad{4\sin\pars{10\c}\sin\pars{40\c} + 1 \over 4\sin\pars{10\c}\sin\pars{40\c} - 1}
={2 \over -2\cot\pars{35\c}\tan\pars{35\c - x}}
\end{align}

$$
\tan\pars{35\c - x}=
{1 - 4\sin\pars{10\c}\sin\pars{40\c} \over 1 + 4\sin\pars{10\c}\sin\pars{40\c}}\,
\tan\pars{35\c}
$$

$$
x = 35\c -\ \underbrace{\arctan\pars{{1 - 4\sin\pars{10\c}\sin\pars{40\c} \over 1 + 4\sin\pars{10\c}\sin\pars{40\c}}\,
\tan\pars{35\c}
}}_{\ds{15\c - n\times 180\c\,,\quad n \in {\mathbb Z}}}
$$

