How do you solve #sin4x= -2sin2x#?

2 Answers
Nghi N.
Jun 5, 2015

f(x) = sin 4x + 2sin 2x = 0

= 2sin 2x.cos 2x + 2sin 2x = 2sin 2x (cos 2x - 1)
Next, solve the 2 basic trig equations: sin 2x = 0 and cos 2x = 1.
a. sin 2x = 0 --> 2x = 0 -> x = 0
sin 2x = 0 -> 2x = pi -> x = pi/2
b. cos 2x = 1 -> 2x = 0 -> x = 0
cos 2x = 1 -> 2x = 2pi -> x = pi.

Cesareo R.
May 12, 2016

#x = k( pi/2)# with #k = {0,1,2,3,...}#

Explanation:

Making #y=2x# and #f(y)=sin(2y)+2sin(y) = 2cos(y)sin(y)+2sin(y)=2(cos(y)-1)sin(y)#.The solutions of #f(y)=0# are the solutions of #cos(y)=1# and #sin(y)=0# which are: #y = 2k pi# and #y = (2k+1)pi#. but #y =2x# so #x = k (pi/2)# with #k = {0,1,2,3,...}#

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