How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = sqrt(2) cosx, 0<=x<=pi?

1 Answer
Nghi N.
Mar 24, 2017

pi/2; pi/4; (3pi)/4

Explanation:

Use trig identity : sin 2x = 2sin x.cos x
Substitute in the equation sin 2x by 2sin x.cos x
2sin x.cos x - sqrt2cos x = 0
cos x(2sin x - sqrt2) = 0
Either one of the 2 factors should be zero:
a. cos x = 0 --> x = pi/2, and x = (3pi)/2
b. 2sin x - sqrt2 = 0
sin x = sqrt2/2
Trig unit circle gives 2 solutions:
x = pi/4 and x = (3pi)/4
Answers for (0, pi)
pi/2; pi/4; (3pi)/4

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