Trigonometric equations questions

$5 \cos^2 (\theta) + 7 \sin^2 (\theta) - 6 = 0$
$5(1- \sin^2 (\theta)) + 7 \sin^2 (\theta) - 6 = 0$
$2 \sin^2 (\theta) =1$
$\sin^2 (\theta) = \frac {1}{2}$
$\sin^2 (\theta) = \sin^2 (\frac {\pi}{4})$
$\theta = n \pi \pm \frac {\pi}{4} , n \in Z$

$\sin^2 (x) = \frac {3}{4}$
$\sin^2 (x) = sin^2 \frac {\pi}{3}$
$x = n \pi \pm \frac {\pi}{3} , n \in Z$

This is of the form
$a \cos (\theta) + b \sin (\theta) = c$
So, $a = \sqrt 3$, b=1 and $c = \sqrt 2$
$\sqrt 3 = r \cos (\alpha)$
$1 = r \sin (\alpha)$
$r=\sqrt {1^2 + (\sqrt 3)^2} = 2$
$\tan (\alpha) = \frac {1}{ \sqrt 3}$
$\alpha = \frac {\pi}{6}$
so, equation becomes
$\sqrt 3 \cos (\theta) + \sin (\theta) = \sqrt 2$
$r \cos (\alpha) \cos (\theta) + r \cos (\alpha) \sin (\theta) = \sqrt 2$
$r \cos( \theta - \alpha)= \sqrt 2$
$2 \cos (\theta - \frac {\pi}{6})=\sqrt 2$
$\cos (\theta - \frac {\pi}{6}) = \frac {1}{\sqrt 2}$
$\cos (\theta - \frac {\pi}{6}) = \cos \frac {\pi}{4}$
$\theta - \frac {\pi}{6}= 2n \pi \pm \frac {\pi}{4}$
$\theta = 2n \pi \pm \frac {\pi}{4} + \frac {\pi}{6}$
$\theta = 2n \pi + \frac {\pi}{4} + \frac {\pi}{6}$ or $\theta = 2n \pi - \frac {\pi}{4} + \frac {\pi}{6}$
$\theta = 2n \pi + \frac {5\pi}{4}$ or $\theta = 2n \pi - \frac {\pi}{12}$

This is of the form
$a \cos (\theta) + b \sin (\theta) = c$
So, $a = \sqrt 3 -1$, $b=\sqrt 3 + 1$ and $c = 2$
$\sqrt 3 -1 = r \cos (\alpha)$
$\sqrt 3 + 1 = r \sin (\alpha)$
$r=\sqrt {(\sqrt 3 -1)^2 + (\sqrt 3 +1)^2} = 2 \sqrt 2$
$\tan (\alpha) = \frac {\sqrt 3 -1}{ \sqrt 3+1}$
$\tan (\alpha) = \tan ( \frac {\pi}{4} - \frac {\pi}{6})$\alpha = \frac {\pi}{12}$so, equation becomes$(\sqrt 3 - 1) \cos (\theta) + (\sqrt 3 + 1) \sin (\theta) = 2 r \cos (\alpha) \cos (\theta) + r \cos (\alpha) \sin (\theta) = 2r \cos( \theta - \alpha)= 22\sqrt 2 \cos (\theta - \frac {\pi}{12})=\sqrt 2\cos (\theta - \frac {\pi}{12}) = \frac {1}{\sqrt 2}\cos (\theta - \frac {\pi}{12}) = \cos (\frac {\pi}{4})\theta - \frac {\pi}{12}= 2n \pi \pm \frac {\pi}{4} \theta = 2n \pi \pm \frac {\pi}{4} + \frac {\pi}{12}$

sin (2x) - sin (4x) + sin (6x) = 0.
or 2 sin (4x) cos(2x) - sin (4x) = 0
i.e. sin (4x)(2 cos(2x) - 1) = 0
Therefore sin (4x) = 0 or 2cos (2x) -1=0
i.e.
sin(4x)= 0 or cos(2x) =1/2
sin(4x)= 0 or $\cos (2x) =\cos (\frac {\pi}{3})$
Hence
$4x=n\pi$ or $2x =2 n \pi \pm \frac {\pi}{3}$
or
$x= \frac {n \pi}{4}$ or $x = n \pi \pm \frac {\pi}{6}$ for $n \in Z$

x= n \pi + (-1)^{n+1} (\frac {3 \pi}{10})$

$-3 \leq a \leq -2$
$x= n \pi \pm cos ^{-1} \sqrt {a+3}$

$\tan (5x) = \cot (2x)$
$\tan (5x)= \tan (\frac {\pi}{2} - 2x)$
or
$5x = n \pi + \frac {\pi}{2} - 2x$
$x = \frac {1}{7} ( n\pi + \frac {\pi}{2})$

$x= (2n+1) \frac {\pi}{10}$