How do I solve the trigonometric equation $\sec^3x - 2 \tan^2 x = 2$?

Question

A friend asked to me how could she resolve this equation, but I don't know how to resolve it?? Could you help me?. The equation is :

$\sec^3x - 2 \tan^2 x = 2$

Note: She told me that I can use the quadratic equation

Answer 1

Or too

$\sec^3x-2\tan^2x=2\\ \frac{1}{\cos^3x}−2 \tan^2 x = 2\\ \frac{1}{\cos^3x}−2 \frac{\sin^2 x}{\cos^2 x} = 2\\ \frac{1}{\cos x}−2 \sin^2 x = 2 \cos^2x\\ \frac{1}{\cos x} = 2 \cos^2x + 2 \sin^2 x = 2(\cos^2x + \sin^2 x) = 2\\ \cos x = \frac{1}{2}\\ x = \cdots$

Answer 2

Multiplying both sides of $$\sec^3x-2\tan^2x=2$$ by $\cos^3x$ gives you $$1-2\sin^2x\cos x=2\cos^3 x$$ $$\iff 1-2(1-\cos^2x)\cos x=2\cos^3x\iff \cos x=1/2.$$ Hence, the answer is $$x=2n\pi\pm \frac{\pi}{3}\ \ (n\in\mathbb Z).$$

Answer 3

I think I have a simpler answer. It seems the easiest way to get this equation to depend on only one trig function is to convert the tangent to a secant yielding

$$\sec^3x-2(\sec^2x-1)=2$$ $$\sec^3x-2\sec^2x+2=2$$ $$\sec^3x-2\sec^2x=0$$

So we have $\sec x=0$ which is impossible or $\sec x=2$. If this doesn't look familiar, take the reciprocal of both sides.