<s>[INST] $15 [\tan 2\theta + \sin 2\theta] + 8 = 0$ if ... 
QUESTION: $15[\tan 2\theta + \sin 2\theta] + 8 = 0$ if...
   (a) $\tan\theta = \frac{1}{2}$
   (b) $\sin\theta = \frac{1}{4}$
   (c) $\tan\theta = 2$
   (d) $\cos\theta = \frac{1}{5}$

What I have tried:
$$\tan2\theta + \sin2\theta = -\frac{8}{15}$$
$$\frac{2\tan\theta}{1-\tan^2\theta} + \frac{2\tan\theta}{1+\tan^2\theta} = -\frac{8}{15}$$
From here onwards, I am stuck and do not know how to continue. Please help. Thank you!
 [/INST] From the final equation you gave, one gets
$$\frac{4\tan\theta}{1-\tan^4\theta} = -\frac{8}{15} \quad\Rightarrow\quad
8\tan^4\theta - 60\tan\theta -8=0.$$
This quartic factors as $4(\tan\theta-2)(2\tan^3\theta + 4\tan^2\theta + 8\tan\theta + 1)$.
This shows that (c) is a solution while (a) is not. To see that (b) and (d) are not solutions, I think you'll have to substitute the appropriate values of $\tan\theta$ into the cubic.
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