sin x = cos x
tan x = 1
x = $\frac{\pi }{4}$

$secxsi{n}^{2}x=tanx\frac{sinx}{cosx}sinx=tanxtanxsinx=tanxtanx=0orsinx=1x=0orx=\frac{\pi }{2}$
But sec x and tan x are not defined at $x=\frac{\pi }{2}$
Therefore, x = 0 is the only solution.

$2co{s}^{2}x-3cosx-2=0$
Let cos x = t
$2t^2-3t-2=0t=\frac{3\pm \sqrt{9-4\times 2\times (-2)}}{2\times 2}t=\frac{3\pm \sqrt{25}}{4}t=2ort=-\frac{1}{2}$
Since cos x cannot be 2, we have
$cosx=-\frac{1}{2}x=\frac{2\pi }{3}$ 

$3se{c}^{2}x=secxsecx=0orsecx=\frac{1}{3}Butsecx\ge 0\forall x$
The equation has no solutions.

sin 2x + cos x=0
2sin x cos x + cos x = 0
cos x(2sin x + 1) = 0
cos x = 0 or sin x = -$\frac{1}{2}$
x = $\frac{\pi }{2}$ or x = -$\frac{\pi }{6}$

$sin2x-sin4x+sin6x=0(sin2x+sin6x)-sin4x=02sin4xcos2x-sin4x=0sin4x(2cos2x-1)=0sin4x=0orcos2x=\frac{1}{2}4x=n\pi or2x=2n\pi \pm \frac{\pi }{3}x=\frac{n\pi }{4}orx=n\pi \pm \frac{\pi }{6}$

The general solution of $sinx=sin\alpha ,\alpha ϵ[-\frac{\pi }{2},\frac{\pi }{2}]$ is
$x=n\pi +(-1{)}^n\alpha ,nϵZ$

$3ta{n}^{2}x-1=0ta{n}^{2}x=\frac{1}{3}tanx=\pm \frac{1}{\sqrt{3}}x=n\pi +\frac{\pi }{6}orn\pi -\frac{\pi }{6}$

The general solution of $cosx=cos\alpha ,\alpha ϵ[0,\pi ]$ is
$x=2n\pi \pm \alpha ,nϵZ$

$3tan\frac{x}{2}+3=03tan\frac{x}{2}=-3tan\frac{x}{2}=-1\frac{x}{2}=n\pi -\frac{\pi }{4}x=2n\pi -\frac{\pi }{2}$