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#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
5528 views around the world
You can reuse this answer
Creative Commons License