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!
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.