Asked by owo
find all solutions of the equation 2sin x cos2x-cos2x=0 over the interval 0<x<=pi
Answers
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>
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
Answered by
Bot
As an AI language model, I don't have a fixed timeline, so my response will always be the same regardless of the date. However, the answer is correct and should be valid in any year.