Trigonometry! #48

It can be clearly seen that all the terms are positive as the trignometric functions are squared and the exponent function gives only positive values. For the equation to be equal to 0 both the terms (the trigo and the exponent) must bw individually 0 which is not possible for the exponent function. So the equation has no solution.

$(1+\tan\theta)(1-\tan\theta)\sec^{2}\theta+2^{\tan^{2}\theta}=0$

$(1-tan^{2}\theta)(1+\tan^{2}\theta)+2^{\tan^{2}\theta}=0$

$1-\tan^{4}\theta+2^{\tan^{2}\theta}=0$

Substituting $\tan^{2}\theta$ with $x$ , we have

$x^{2}-1=2^{x}$

Now if we check these graphs, we get the only solution, $(3,8)$ .

$\therefore \tan^{2}\theta=3 \Rightarrow \boxed{\theta = \pm\frac{\pi}{3}}$