SOLUTION: Solve. In radians on the interval [0,2pi). 3sec x = 3 + tan^2 x

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Question 678469: Solve. In radians on the interval [0,2pi).
3sec x = 3 + tan^2 x
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve. In radians on the interval [0,2pi).
3sec x = 3 + tan^2 x
3/cosx=3+sin^2x/cos^2x
LCD:cos^2x
3cosx=3cos^2x+sin^2x
3cosx=3cos^2x+(1-cos^2x)
3cosx=3cos^2x+1-cos^2x
2cos^2x-3cosx+1=0
(2cosx-1)(cosx-1)=0
2cosx-1=0
cosx=1/2
x=π/3, 5π/3
..
cosx-1=0
cosx=1
x=0
..
x=0, π/3, 5π/3