The principle solutions lies in the range 0 ≤ x ≤ 2π. We can principle solution using the below method
i. Find the value of the angle using the positive value.
ii Then remember these below identities to find the principle values
$\sin( \pi -x )=\sin x$
$\sin ( \pi + x) = -\sinx $
$\sin ( 2\pi - x) = -\sinx $
$\cos ( \pi -x )= -\cos x$
$\cos ( \pi +x )= -\cos x$
$\cos ( 2\pi -x )= \cos x$
$\tan ( \pi -x )= -\tan x$
$\tan ( \pi +x )= \tan x$
$\tan ( 2\pi -x )= -\tan x$
(a) $\sin (x)= \frac {\sqrt 3}{2}$
$\sin (x)= \sin (\pi/3)$
Now $sin (\pi - \pi/3) = sin \pi/3$
or $ sin (2 \pi/3) = \frac {\sqrt 3}{2}$
So principle solutions are π/3, 2π/3
(b) $\cos (x) =- \frac {1}{\sqrt 2}$
$\cos (x) = - \cos (\pi /4)$
Now $\cos (\pi - \pi /4) = -\cos (\pi /4)$
or $\cos (3 \pi/4) = - \frac {1}{\sqrt 2}$
Also $\cos (\pi + \pi /4)= - \cos (\pi /4)$
or $\cos (5 \pi/4) = - \frac {1}{\sqrt 2}$
So principle solutions are 3π/4, 5π/4
(c) tanx =-1
$ \tan (x) = - \tan (\pi/4)$
Now $ \tan ( \pi -\pi/4 )= -\tan (\pi/4)$
or $ \tan (3\pi/4) = -1$
Also $\tan ( 2\pi - \pi/4 )= -\tan \pi/4$
or $ \tan (7 \pi/4) = -1$
So principle solutions are 3π/4, 7π/4
$ \sin (\frac {\theta }{2}) =-1$
$\sin (\frac {\theta }{2})= \sin ( -\frac {\pi}{2})$
$\frac {\theta }{2= n \pi + (-1)^n ( -\frac {\pi}{2}) ,n \in Z$
$\theta = 2 n \pi + (-1)^{n+1} \pi ,n \in Z$
$ \cos (3\theta) = - \frac {1}{2} $
$\cos (3\theta) =\cos (\frac {2 pi}{3})$
$3 \theta= 2n \pi \pm \frac {2 pi}{3} , n \in Z$
$\theta= \frac {2n \pi}{3} \pm \frac {2 pi}{9} , n \in Z$
$5 \cos^2 (\theta) + 7 \sin^2 (\theta) - 6 = 0$
$5(1- \sin^2 (\theta)) + 7 \sin^2 (\theta) - 6 = 0$
$2 \sin^2 (\theta) =1$
$\sin^2 (\theta) = \frac {1}{2}$
$\sin^2 (\theta) = \sin^2 (\frac {\pi}{4})$
$\theta = n \pi \pm \frac {\pi}{4} , n \in Z$
$ \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) = 2$
$r \cos( \theta - \alpha)= 2$
$2\sqrt 2 \cos (\theta - \frac {\pi}{12})=\sqrt 2$
$\cos (\theta - \frac {\pi}{12}) = \frac {1}{\sqrt 2}$
$\cos (\theta - \frac {\pi}{12}) = \cos (\frac {\pi}{4})$
$\theta - \frac {\pi}{12}= 2n \pi \pm \frac {\pi}{4}$
$ \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
$5x = n \pi + \frac {\pi}{2} - 2x$
$x = \frac {1}{7} ( n\pi + \frac {\pi}{2})$
$x= (2n+1) \frac {\pi}{10}$
