<s>[INST] To Substitute or Not ... $2\sin 2x = 3\tan x$, in interval $0\le x<360^\circ$ The Question
Solve $2\sin 2x = 3\tan x$, in interval $0\le x<360^\circ$
I thought this straightforward enough, used the identity $\sin^2 x + \cos^2 x \equiv 1$, shown below in the working, which clearly is not right. My question is what have I done wrong here - why is the substitution of $(1-\sin^2x)$ for $\cos^2x$ wrong?
My Solution
$$\Rightarrow 4\sin x \cos x = \frac{3\sin x}{\cos x}$$
$$\Rightarrow 4\sin x \cos^2 x = 3\sin x$$
$$\Rightarrow 4\sin x(1 - \sin^2 x) - 3\sin x = 0$$
$$\Rightarrow \sin x - 4\sin^3x = 0$$
$$\Rightarrow \sin x(1 - 4\sin^2 x) = 0$$
$$\Rightarrow \sin x = 0, \sin^2 x = \frac{1}{4}$$
The alarm bells started ringing at this point as $2\sin 90^\circ \neq 3\tan 45^\circ$
Correct Approach
So, looking at it, using the identity $\sin^2 x + \cos^2 x \equiv 1$ was the error.
So:
$$\Rightarrow 4\sin x \cos x = \frac{3\sin x}{\cos x}$$
$$\Rightarrow 4\sin x \cos^2 x = 3\sin x$$
$$\Rightarrow \sin x (4\cos^2 x - 3) = 0$$
$$\Rightarrow \sin x = 0, \cos^2 x = \frac{3}{4}$$
I'd be very grateful if anyone could explain why making the substitution I made was wrong. I thought they could be used as-and-when. Have I misunderstood something significant?
Or is is that the substitution can't be used because it gives the wrong answer? But this begs the question of how can you know when it's OK to use it.
Thanks in advance for your insights.
 [/INST] Your original solution is correct as well. It's just that $x=45^\circ$ doesn't satisfy $\sin^2 x = \frac14$. If $\sin^2 x = \frac14$, then $\sin x = \pm\frac12$, while $\sin 45^\circ = \frac{1}{\sqrt{2}}$. The solutions for $\sin^2 x = \frac14$ with $0\leq x<360^\circ$ are $30^\circ, 150^\circ, 210^\circ$ and $330^\circ$.
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