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

1.	Solve 2 tan2 x + sec2 x = 2 for 0 < x < 2π
Answer» 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

Answer»

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

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