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 Solution
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
Question 8.
Solve sin (2x) - sin (4x) + sin (6x) = 0. Solution
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 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 Solution
$-3 \leq a \leq -2$
$x= n \pi \pm cos ^{-1} \sqrt {a+3}$