Finding the general solution for complex trig equations.

Question

• $4\tan^2 x-3=0$

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• $ \tan^2 x =\frac34$
• $x = \tan^{-1}(\pm \frac{\sqrt{3}}{2}$)

therefore general solution

$x= \tan^{-1}(\pm \frac{\sqrt{3}}{2}) + n\pi$ where $n$ is an element of all real numbers.

Answer 1

Basics: The set of solutions to the equation $\tan x=a$ is given by $$\left\{ \tan^{-1}(a)+n\pi\;\left|\right.\; n\in\mathbb Z \right\}.$$ So in your particular example: $$\eqalign{ 4\tan^2 x-3=0 & \iff \tan^2x=\dfrac34 \\ &\iff \tan x=\pm\dfrac{\sqrt{3}}2 \\ &\iff x=\tan^{-1}\left(\pm\dfrac{\sqrt{3}}2\right)+n\pi.\qquad n\in\mathbb Z \\ }$$ So your error is that you considered $n$ to be a real number, but that's false, since then the set of solutions would have been the entire $\mathbb R$.