Solving for x in the given interval NEED HELP !!!

Hello, math71321!

Quote:

$b)\;\;\cos^2\!x + 2\sin x\cos x-\sin^2\!x\:=\:0 \qquad0 \leq x \leq 2\pi$

We have: . $\underbrace{\cos^2\!x - \sin^2\!x} + \underbrace{2\sin x\cos x} \;=\;0$
. . . . . . . . . . $\cos2x \quad\;\;+ \quad\;\;\sin 2x \quad=\;0$

Then: . $\sin2x \:=\:-\cos2x \quad\Rightarrow\quad\frac{\sin2x}{\cos2x} \:=\:-1 \quad\Rightarrow\quad \tan2x \:=\:-1$

Hence: . $2x \;=\;\frac{3\pi}{4},\:\frac{7\pi}{4},\:\frac{11\pi }{4},\:\frac{15\pi}{4}$

Therefore: . $\boxed{x \;=\;\frac{3\pi}{8},\:\frac{7\pi}{8},\:\frac{11\pi }{8},\:\frac{15\pi}{8}}$

Quote:

$c)\;\;2\tan x \:= \:\sec x\qquad -\pi \leq x \leq 2\pi$

Square both sides: . $4\tan^2\!x \:=\:\sec^2\!x$
. . . . . . . . . . . . $4\overbrace{(\sec^2\!x - 1)} \:=\:\sec^2\!x$

which simplifies to: . $3\sec^2\!x \:=\:4\quad\Rightarrow\quad\sec^2\!x \:=\:\frac{4}{3}\quad\Rightarrow\quad \sec x \:=\:\pm\frac{2}{\sqrt{3}}$

. . Therefore: . $\boxed{x \;=\;\pm\frac{\pi}{6},\:\pm\frac{5\pi}{6},\:\frac{ 7\pi}{6},\:\frac{11\pi}{6}}$