Solve the equation 2cos^2x=sin2x for 0 is < or= x is < or = pi

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Question: Solve the equation 2cos^2x=sin2x, for 0 <= x <= pi

2cos^2x = sin2x

2cos^2x = 2sinx.cosx

cos^2x - sinx.cosx = 0

cosx(cosx - sinx) = 0

Hence, cosx = 0,  cosx - sinx = 0 => tanx = 1

So, x = pi/2, 3pi/2 , x = pi/4, 5pi/4

by Level 11 User (81.5k points)

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