# Trigonometric equations questions

Question 1.
Find the Principal solutions of the below trigonometric equations
a. $\sin (x)= \frac {\sqrt 3}{2}$
b. $\cos (x) =- \frac {1}{\sqrt 2}$
c. tan x=-1

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

Find the general solution of the equation
Question 2.
$\sin (\frac {\theta }{2}) =-1$

$\sin (\frac {\theta }{2}) =-1$
$\sin (\frac {\theta }{2})= \sin ( -\frac {\pi}{2})$
$\frac {\theta }{2= n \pi + (-1)^n ( -\frac {\pi}{2}) ,n \in Z$
$\theta = 2 n \pi + (-1)^{n+1} \pi ,n \in Z$

Question 3.
$\cos (3\theta) = - \frac {1}{2}$

$\cos (3\theta) = - \frac {1}{2}$
$\cos (3\theta) =\cos (\frac {2 pi}{3})$
$3 \theta= 2n \pi \pm \frac {2 pi}{3} , n \in Z$
$\theta= \frac {2n \pi}{3} \pm \frac {2 pi}{9} , n \in Z$

Question 4.
$5 \cos^2 (\theta) + 7 \sin^2 (\theta) - 6 = 0$

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

Question 5.
$\sin^2 (x) = \frac {3}{4}$

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

Question 6.
$\sqrt 3 \cos (\theta) + \sin (\theta) = \sqrt 2$

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 $\theta = 2n \pi - \frac {\pi}{12}$

Question 7.
Find the general solution of the equation
$(\sqrt 3 - 1) \cos (\theta) + (\sqrt 3 + 1) \sin (\theta) = 2$

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}$Question 8. Solve sin (2x) - sin (4x) + sin (6x) = 0. Answer sin (2x) - sin (4x) + sin (6x) = 0. or 2 sin (4x) cos(2x) - sin (4x) = 0 i.e. sin (4x)(2 cos(2x) - 1) = 0 Therefore sin (4x) = 0 or 2cos (2x) -1=0 i.e. sin(4x)= 0 or cos(2x) =1/2 sin(4x)= 0 or$\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$n \in Z$Question 9.$4 \cos^2 (x) \sin (x) - 2 \sin^2 (x) = 3 \sin (x)$Answer x= n \pi + (-1)^{n+1} (\frac {3 \pi}{10})$

Question 10.
Determine all the values of b for which the equation $\cos^4 (x) - (b+2) \cos^2 (x) -(b+3)=0$ possesses solution . Find the solutions also

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

Question 11.
Solve the below equation
$\tan (5x) = \cot (2x)$

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

Question 12.
Solve the below equation
$\sec (4x) - \sec(2x) =2$

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