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August 28, 2014

August 28, 2014

Posted by **Help** on Tuesday, July 2, 2013 at 5:10am.

2 (sec x)^2 + (tan x)^2 -3 = 0

- Pre-cal -
**Steve**, Tuesday, July 2, 2013 at 9:44amsince sec^2 x = 1+tan^2 x, we have

2(1+tan^2 x)+tan^2 x-3=0

3tan^2 x-1 = 0

tan^2 x = 1/3

tan x = ±1/√3

tan x > 0 in QI, QIII, so

x = π/6, 7π/6

tan x < 0 in QII, QIV, so

x = 5π/6, 11π/6

Actually, since the period of tan x is

π, it would be better to ask for solutions in [0,π) rather than 2π.

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