Question

Solve 2 tan² x + sec² x = 2 for 0 < x < 2π

Answer 1

We have: 2 tan² x+ sec² x= 2 

2 tan² x + 1 + tan² x = 2 

Which gives tan x = ± \(\frac{1}{\sqrt{3}}\) 

If we take tan x = \(\frac{1}{\sqrt{3}}\) ,then x = \(\frac{\pi}{6}\) or \(\frac{7\pi}{6}\) 

Again, if we take tanx = −\(\frac{1}{\sqrt{3}}\) , then x = \(\frac{5\pi}{6}\) or  \(\frac{11\pi}{6}\) 

Therefore, the possible solutions of above equations are 

x = \(\frac{\pi}{6}\) , \(\frac{5\pi}{6}\) , \(\frac{7\pi}{6}\)  and \(\frac{11\pi}{6}\) Where 0 ≤ x ≤ 2π