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

Question

asked Aug 27, 2020 13.0k views

1 Answer

RePierre · Answer 1 · 2020-09-01T01:55:00+0000

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

$3\tan^2x-2=\csc^2x-\cot^2x\iff3(\sin^2x)/(\cos^2x)-2=\frac1{\sin^2x}-(\cos^2x)/(\sin^2x)$

Turn everything into fractions with common denominators:

$(3\sin^2x-2\cos^2x)/(\cos^2x)=(1-\cos^2x)/(\sin^2x)$

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

On the left:

$(3\sin^2x+3\cos^2x-5\cos^2x)/(\cos^2x)=(3-5\cos^2x)/(\cos^2x)$

On the right:

$(1-\cos^2x)/(\sin^2x)=(\sin^2x)/(\sin^2x)=1$

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

So we have

$(3-5\cos^2x)/(\cos^2x)=1\implies3-5\cos^2x=\cos^2x$

$\implies3=6\cos^2x$

$\implies\cos^2x=\frac12$

$\implies\cos x=\pm\frac1{\sqrt2}$

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