Question 181755: Solve the equation for the interval [0,2л):
tan2x-2cosx=0
PLEASE HELP ME!!!!!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! tan(2x)-2cos(x)=0
2(tanx/(1-tan^2(x))) - 2cosx = 0 (Identity from Google)
(tan(x))/(1-tan^2(x)) - cos(x) = 0
tan - cos(1-tan^2) = 0 All arguments x
tan - cosx + sin^2/cos = 0
sin - cos^2 + sin^2 = 0
sin + (cos^ + sin^2) = 2cos^2
sin + 1 = 2 - 2sin^2
2sin^2 + sin - 1 = 0 Quadratic in sin(x)
sin = (-1 +- sqrt(9))/4
sin(x) = 1/2, -1
x = pi/6, 3pi/2, 5pi/6
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