Question

Solve this sec^2=4/3

Answer 1

$\sec ^{2} ( \theta) = \frac{4}{3}$

$\implies \sec( \theta) = \frac{ 2 }{ \sqrt{3} }$

$\implies \cos( \theta) = \frac{ \sqrt{3} }{2} \:\:\: \: ( \theta \not = \frac{(2n + 1)\pi}{2} ,n \in \mathbb{Z})$

$\implies \cos( \theta )= \cos( \pm \frac{\pi}{6} + n\pi )$

$\implies \boxed{\theta = \pm \frac{\pi}{6} + n\pi}\: \: \: \: \: \: \: \: \: ( n \in \mathbb{Z})$