<s>[INST] Trigonometric identity for $a\sin 2x + b\sin(x+\alpha )$ The following is a known trigonometric identity,
$$a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,$$
where
$$c=\sqrt {a^2+b^2+2ab\cos \alpha },\,$$
and
$$\beta =\arctan \left(\frac {b\sin \alpha }{a+b\cos \alpha }\right) + \begin{cases} 0 & \text{if } a+b \cos \alpha \geq 0,\\
\pi & \text{if } a+b\cos \alpha <0.\end{cases}$$
Is there an identity for the following ?
$$a\sin 2x + b\sin(x+\alpha )$$
Alternatively, can you find the values of $x$ that satisfy the following equation?
$$a\sin 2x - b\sin(x+\alpha )=0$$
 [/INST] As for the last question:
\begin{align*}
a\sin2x-b\sin(x+\alpha)&=0 \\
2a\sin x\cos x - b(\sin x\cos\alpha + \cos x\sin\alpha)&=0\\
-b\cos\alpha\sin x &= \cos x \cdot(b\sin\alpha-2a\sin x) \\
\frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} &= \cos x \\
\bigg( \frac{-b\cos\alpha\sin x}{b\sin\alpha-2a\sin x} \bigg)^2 &= \cos^2 x = 1-\sin^2x
\end{align*}
gives you an equation of degree $4$ in $\sin x$ that you can try to solve your favorite way. (You'll need to check any solutions against the original equation, since squaring both sides will introduce extraneous solutions.)
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