How do you solve #sin2x+sin4x=cos2x+cos4x# for x in the interval [0,2pi)?

1 Answer
Mar 18, 2016

#pi/12 and (5pi)/12#

Explanation:

Apply the sum- to-product identities:
#sin a + sin b = 2sin ((a + b)/2).cos ((a -b)/2)#.
In this case:
#sin 2x + sin 4x = 2sin ((2x + 4x)/2).cos ((4x - 2x)/2) =#
#= 2sin (3x).cos x# (1)
#cos 2x + cos 4x = 2cos ((2x + 4x)/2).cos ((4x - 2x)/2) =#
#= 2cos (3x).cos x.#(2)
We have : (1) = (2)
2sin (3x).cos x = 2cos (3x).cos x
After simplification:
sin (3x) = cos (3x)
Divide both sides by cos 3x
#tan 3x = 1 = tan (pi/4)# -->
Trig table and unit circle give 2 solutions:
a. #3x = pi/4# --> #x = pi/12.#and
b. #3x = pi/4 + pi = 5pi/4 #--> #x = (5pi)/12#
Check.
#x = pi/12 = 180/12 = 15^@# --> #2x = 30^@# --> #4x = 60^@#
#sin 30 + sin 60 = 1/2 + sqrt3/2#
#cos 30 + cos 60 = sqrt3/2 + 1/2#. Proved