How do you solve $2cos2x=1-cosx$ for $0<=x<=360$ ?

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Answer 1 · 2016-10-30T13:52:45

Apply the identity $cos2x = 1- 2sin^2x$ .

$2(1 - 2sin^2x) = 1 - cosx$

$2 - 4sin^2x = 1 - cosx$

$2 - 4(1 - cos^2x) = 1 - cosx$

$2 - 4 + 4cos^2x = 1 - cosx$

$4cos^2x + cosx - 3 = 0$

Let $t = cosx$ .

$4t^2 + t - 3 = 0$

$4t^2 + 4t - 3t - 3 = 0$

$4t(t + 1) - 3(t + 1) = 0$

$(4t - 3)(t + 1) =0$

$t = 3/4" AND " -1$

$cosx = 3/4" AND " cosx = -1$

$x = arccos(3/4), 360 - arccos(3/4)" AND "x = 180˚$

Hopefully this helps!

How do you solve 2cos2x=1-cosx2cos2x=1-cosx for 0<=x<=3600<=x<=360?