How do you solve cos2x = sinxcos2x=sinx on the interval 0 <= x<= 2pi0≤x≤2π?

1 Answer
Alan P.
Apr 16, 2015

If cos(2x) = sin(x)cos(2x)=sin(x)
then
1-2sin^2(x) = sin(x)1−2sin2(x)=sin(x)

2sin^2(x) +sin(x) -1 =02sin2(x)+sin(x)−1=0

Substituting k=sin(x)k=sin(x)
2k^2+k-1 = 02k2+k−1=0

(2k-1)(k+1) = 0(2k−1)(k+1)=0

sin(x) = 1/2sin(x)=12 or sin(x) =-1sin(x)=−1

If sin(x) = 1/2sin(x)=12 (for 0<=x<=2pi0≤x≤2π)
x=pi/6 =30^ox=π6=30o or x= (5pi)/6 = 150^ox=5π6=150o

If sin(x) = -1sin(x)=−1 (for 0<=x<=2pi0≤x≤2π)
x=(3pi)/2 = 270^ox=3π2=270o

So x epsilon {pi/6, (5pi)/6, (3pi)/2}xε{π6,5π6,3π2}
(or their equivalent in degrees)

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