What is the solution to #tan^2x= 3# on #[0, 2pi)#?

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Answer 1 · 2017-11-17T18:30:34

#x = pi/3, (2pi)/3, (4pi)/3, (5pi)/3#

We can rewrite as

#sin^2x/cos^2x = 3#

#sqrt(sin^2x/cos^2x) = +-sqrt(3)#

#sinx/cosx = +-sqrt(3)#

#tanx= +-sqrt(3)#

Taking #arctan#, we have:

#x = pi/3 and ?#

We take the three other angles that have #pi/3# for a reference angle.

#x = pi/3, (2pi)/3, (4pi)/3, (5pi)/3#

Hopefully this helps!