Question

Find the principal solution of
(i) sin x =-root 3/2
(ii) cos x =-1/2
(iii) cot x =-root 3

Answer 1

Hi ....Nidhi.
Here is your answer....

Answer 2

Some of the general formulas,

If, $\sin( \theta) = \sin( \alpha )$

then, $\theta = n\pi + ( { - 1})^{n} \alpha$

If, $\cos( \theta) = \cos( \alpha )$

then, $\theta = 2n\pi \pm \alpha$

Solution lying in the range of [ 0, 360°] are called Principal solutions.

____________________________________________________________________

$1) \sin(x) = - \frac{ \sqrt{3} }{2} \\ \\ \sin(x) = - sin( \frac{\pi}{3} ) \\ \\ \sin(x) = \sin( - \frac{\pi}{3} ) \\ \\ x = n\pi + ( - 1) ^{n} ( \frac{ - \pi}{3} )$

For n = 1,

$x = \pi + \frac{ \pi}{3} = \frac{4\pi}{3}$

For n =2,

$x =2 \pi + \frac{ - \pi}{3} = \frac{5\pi}{3}$

Therefore, The principal solutions of

$\sin(x) = \frac{ - \sqrt{3} }{2} \: \sf \: are \: \: \: \frac{4\pi}{3} , \: \frac{5\pi}{3}$

___________________________________________________________________

$2) \cos(x) = - \frac{1}{2} \\ \\ \cos(x) = \cos(\pi - \frac{\pi}{3} ) \\ \\ \cos(x) = \cos( \frac{2\pi}{3} ) \\ \\ x = 2n\pi \pm \: \frac{2\pi}{3}$

For n = 0,

$x = \pm \frac{2\pi}{3}$

For n =1,

$x = 2\pi \pm \: \frac{2\pi}{3}$

Therefore, The principal solutions of

$\cos(x) = \frac{ - 1 }{2} \: \sf \: are \: \: \: \frac{2\pi}{3}, \: \frac{4\pi}{3}$

__________________________________________________________________

$3) cotx = - \sqrt{3} \\ \\ \tan(x) = - \frac{1}{ \sqrt{3} } \\ \\ \tan(x) = - \tan( \frac{\pi}{6} ) \\ \\ \tan(x) = \tan( - \frac{\pi}{6} ) \\ \\ x = n\pi - \frac{\pi}{6}$

For n = 1,

$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

For n = 2,

$x =2 \pi - \frac{\pi}{6} = \frac{11\pi}{6}$

Therefore, The principal solutions of

$\cot(x) = - \sqrt{3} \: \sf \: are \: \: \: \frac{5\pi}{6}, \: \frac{11\pi}{6}$

________________________________________________________________