Asked by owo

find all solutions of the equation 2sin x cos2x-cos2x=0 over the interval 0<x<=pi
Answered by Reiny
2sin x cos2x-cos2x=0
cos2x(2sinx - 1) = 0

cos 2x = 0
2x = 90° , 270°, 450, 630°
x = 45° , 135°, 225°, 315° OR in radians: <b>π/4, 3π/4, 5π/4, 7π/4</b>

2sinx - 1 = 0
sinx = 1/2
x = 30°, 150° OR in radians: <b>π/6, 5π/6</b>
Answered by Sunwoo
B. x = pi/6, pi/4, 3pi/4, 5pi/6

Answered by kaz
still correct as of 2023
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