Question #7fce1

2 Answers
Dec 7, 2017

#x = pi/4 + 2pik, pi/2 + 2pik, 3pi/4 + 2pik# (Period of sin(x) and cos(x) is #2pi#)

Explanation:

Assuming #sin(2x)sin(x) = cos(x)#, begin by using the double angle identity, #sin(2u) = 2sin(u)cos(u)# to rewrite the equation.

#sin(2x)sin(x) = cos(x) rarr 2sin(x)sin(x)cos(x) = cos(x) rarr 2sin^2(x)cos(x) = cos(x)#

Next subtract cos(x) from the right side to the left to make the equation equal to zero.

#2sin^2(x)cos(x)-cos(x) = 0#

Factor out cos(x) and separate

#cos(x)*(2sin^2(x)-1) = 0 rarr cos(x) = 0# and #2sin^2(x)-1 = 0#

Find the exact values using the unit circle or right triangle method, #cos(x) = 0# when #x = pi/2# and #sin(x) = sqrt(2)/2# when #x = pi/4, 3pi/4# (Remember restrictions #sin(x): R(-pi/2, pi/2); cos(x): R(0, pi)#

Dec 7, 2017

Answers:
#x = pi/2 + kpi#
#x = pi/4 + (kpi)/2#

Explanation:

#sin 2x.sin x - cos x = 0#
Replace sin 2x by (2sin x.cos x).
#2sin^2 x.cos x - cos x = 0#
#cos x(2sin^2 x - 1) = 0#
Either factor should be zero.
A. cos x = 0. Unit circle give 2 solutions:
#x = pi/2,# and #x = (3pi)/2#
General answer: #x = pi/2 + kpi #
B. #2sin^2 x - 1 = 0# --> #sin^2 x = 1/2#
#sin x = +- sqrt2/2#
a. #sin x = sqrt2/2#. Trig table and unit circle give 2 solutions:
#x = pi/4 #, and #x = pi - pi/4 = (3pi)/4#
b. #sin x = -sqrt2/2#. Unit circle gives -->
#x = pi + pi/4 = (5pi)/4# and #x = 2pi - pi/4 = (7pi)/4#
General answer: #x = pi/4 + (kpi)/2#