Question

Square roots in trigonometry. I don’t understand please help?

Answer 1

By definitions of the (co)tangent and cosecant function,

$3\tan^2x-2=\csc^2x-\cot^2x\iff3\dfrac{\sin^2x}{\cos^2x}-2=\dfrac1{\sin^2x}-\dfrac{\cos^2x}{\sin^2x}$

Turn everything into fractions with common denominators:

$\dfrac{3\sin^2x-2\cos^2x}{\cos^2x}=\dfrac{1-\cos^2x}{\sin^2x}$

Recall that $\cos^2x+\sin^2x=1$, so we can simplify both sides a bit.

On the left:

$\dfrac{3\sin^2x+3\cos^2x-5\cos^2x}{\cos^2x}=\dfrac{3-5\cos^2x}{\cos^2x}$

On the right:

$\dfrac{1-\cos^2x}{\sin^2x}=\dfrac{\sin^2x}{\sin^2x}=1$

(as long as $\sin x\neq 0$, which happens in the interval $0\le x\le\pi$ when $x=0$ or $x=\pi$)

So we have

$\dfrac{3-5\cos^2x}{\cos^2x}=1\implies3-5\cos^2x=\cos^2x$

$\implies3=6\cos^2x$

$\implies\cos^2x=\dfrac12$

$\implies\cos x=\pm\dfrac1{\sqrt2}$

$\implies x=\dfrac\pi4\text{ or }x=\dfrac{3\pi}4$