<s>[INST] Solving for $\lambda$ and $\mu$ in $5\sin(x) = \lambda(2\sin(x)-3) + \mu 2\cos(x)$ 
Solve for $\lambda$ and $\mu$ in the following equation:
$$5\sin(x) = \lambda(2\sin(x)-3) + \mu 2\cos(x)$$

I have tried:
$$
 5\sin(x) = 2\lambda \sin(x) -3\lambda+2\mu \cos(x)
$$
Comparing the coefficient of $\sin x$:
$$
 2\lambda = 5
$$
$$
 \lambda = \frac{5}{2}
$$
 [/INST] It's $$(2\lambda-5)\sin{x}+2\mu\cos{x}=3\lambda$$ or since the case $2\lambda-5=\mu=0$ is impossible,
$$\frac{2\lambda-5}{\sqrt{(2\lambda-5)^2+4\mu^2}}\sin{x}+\frac{2\mu}{\sqrt{(2\lambda-5)^2+4\mu^2}}\cos{x}=\frac{3\lambda}{\sqrt{(2\lambda-5)^2+4\mu^2}}.$$
Now, take $$\frac{2\lambda-5}{\sqrt{(2\lambda-5)^2+4\mu^2}}=\cos\alpha$$ and
$$\frac{2\mu}{\sqrt{(2\lambda-5)^2+4\mu^2}}=\sin\alpha.$$
Can you end it now?
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