JOMO 5, Long 2

$\sin(x) + \cos(x) + \tan(x) = 0 \Rightarrow \sin(x) + \cos(x) = -\tan(x)$

$\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = \tan^2(x)$

$1 + \sin(2x) = \tan^2(x)$

$\sin(2x) = \frac{\sin^2(x) - \cos^2(x) }{\cos^2(x)}$

$\sin(2x) = \frac{-\cos(2x)}{\cos^2(x)}$

$\tan(2x) = -\sec^2(x)$

$\frac{-2\tan(x)}{1-\tan^2(x)} = \left(1 + \tan^2(x) \right)$

$-2\tan(x) = \left(1- \tan^2(x) \right) \left(1+ \tan^2(x) \right)$

$-2\tan(x) = 1 - \tan^4(x)$

Let $y = \tan(x) \Rightarrow y^4-2y-1 = 0$

$y = 1 \Rightarrow y^4 - 2y - 1 = -2 < 0$

$y = 2 \Rightarrow y^4 - 2y - 1 = 11 > 0$

$y^4 - 2y - 1$ is a continuous function and thus by the Intermediate Value Theorem, there exists some real $a, \; 1 < a < 2$ such that $y = a \Rightarrow y^4 - 2y - 1 = 0$

This then means that since $a \in \mathbb{R}$ then there exists some $x \in \mathbb{R}$ such that $\tan(x) = a$

However, it remains to show that this result is not spurious as the second step involved squaring, and this is something I have yet to accomplish. I shall leave this here and then edit in future as appropriate.

Hmm I have a clear path to a solution which will work and give me an answer, thus solving the problem. The algebra involved is messy and complex and I think there is a much more elegant way to prove that there is a solution without proof by example.