How do you solve cosx=sin2xsinx?

2 Answers
CW
Sep 12, 2016

x=pi/4 or 45^@

Explanation:

recall that sin2x=2sinxcosx

cosx=sin2xsinx
cosx=2sinxcosxsinx
1=2sin^2x
sin^2x=1/2

sinx=1/sqrt2

x=45^@

Nghi N.
Sep 12, 2016

pi/2; (3pi)/2; pi/4; (3pi)/4; (5pi)/4; (7pi)/4

Explanation:

cos x = sin 2x.sin x
cos x = 2sin x.cos x.sin x = 2cos x.sin^2 x
cos x - 2cos x.sin^2 x = 0
Put cos x in common factor:
cos x(1 - 2sin^2 x) = 0
Use trig table, and unit circle to solve this trig equation
a. cos x = 0 --> x = pi/2 and x = (3pi)/2
b. (1 - 2sin^2 x) = 0
sin^2 x = 1/2
sin x = +- 1/sqrt2 = +- sqrt2/2
c. sin x = sqrt2/2 --> x = pi/4 and x = (3pi)/4
d. sin x = - sqrt2/2 --> x = (5pi)/4 and x = (7pi)/4

Answers for (0, 2pi):
pi/2; (3pi)/2; pi/4; (3pi)/4; (5pi)/5; (7pi)/4.
For general answers, add 2kpi

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