If $x$ is a solution of the equation: $$\tan^3 x = \cos^2 x - \sin^2 x$$ Then what is the value of $\tan^2 x$?
This is the problem you are supposed to do it just with highschool trigonometry , but i can't manage to do it please help
Here are the possible answers: $$a) \sqrt{2}-1, b) \sqrt{2}+1, c) \sqrt{3}-1, d) \sqrt{3}+1, e)\sqrt{2}+3$$
Assuming the Left Hand Side to be $\tan^2x$
$$\cos^2x-\sin^2x=\frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$
If we put $\tan^2x=t,$ the equation becomes $$t=\frac{1-t}{1+t}\implies t^2+2t-1=0 $$
$\displaystyle \tan^2x=t=\frac{-2\pm\sqrt{2^2-4\cdot1\cdot(-1)}}2=-1\pm\sqrt2$
If $x$ is real, $t=\tan^2x\ge0$ and we know $\sqrt2>1$
$$\sqrt{t^3} = \frac{1-t}{1+t}$$
For this equation Wolfram Alpha says that $t\approx 0.489632$, which obviously isn't in the possible answers
– Stefan4024 Sep 07 '13 at 14:39Hint: $$(1+\tan^2 x)^2 (\tan^2 x)^3 = \left((1+\tan^2 x) (\tan^3 x)\right)^2 = \left( \frac{1}{\cos^2 x} (\cos^2 x - \sin^2 x) \right)^2 = \left( 1 - \tan^2 x\right)^2.$$
You must have a typo in the question. You can verify that a) the solution involves roots of high degree polynomials, and that b) numerically, none of the solutions (which are very likely not a simple sum of square roots) match your possible answers. For reference, the three first positive solutions are: $$x=0.610549, \ 3.75214, \ 6.89373\ldots$$
$\tan^2 x=\cos^2 x-\sin^2 x$
$\sin^2 x=\cos^4 x -\sin^2 x \cos^2 x$
$0=\cos^4 x -\sin^2 x \cos^2 x -\sin^2 x$
$0=\cos^4 x - \sin^2 x(1+\cos^2 x)$
$0=\cos^4 x - (1-\cos^2 x)(1+\cos^2 x)$
$0=\cos^4 x -(1- \cos^4 x)$
$0=2 \cos^4 x -1$
$\cos^2 x=\frac {\sqrt 2}{2}$
$\large \frac {1}{\cos^2 x}=\sqrt 2$
$\tan^2 x=\sqrt 2 \sin^2 x$
Looks like we're stuck, but from above we have $\cos^2 x=\frac {\sqrt 2}{2}$ , so $\sin^2 x=1-\frac {\sqrt 2}{2}$
By substitution, $\tan^2 x=\sqrt 2 (1-\frac {\sqrt 2}{2})$
$\tan^2 x=\sqrt 2 -1$
