Apply the sum- to-product identities:
sin a + sin b = 2sin ((a + b)/2).cos ((a -b)/2)sina+sinb=2sin(a+b2).cos(a−b2).
In this case:
sin 2x + sin 4x = 2sin ((2x + 4x)/2).cos ((4x - 2x)/2) =sin2x+sin4x=2sin(2x+4x2).cos(4x−2x2)=
= 2sin (3x).cos x=2sin(3x).cosx (1)
cos 2x + cos 4x = 2cos ((2x + 4x)/2).cos ((4x - 2x)/2) =cos2x+cos4x=2cos(2x+4x2).cos(4x−2x2)=
= 2cos (3x).cos x.=2cos(3x).cosx.(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)tan3x=1=tan(π4) -->
Trig table and unit circle give 2 solutions:
a. 3x = pi/43x=π4 --> x = pi/12.x=π12.and
b. 3x = pi/4 + pi = 5pi/4 3x=π4+π=5π4--> x = (5pi)/12x=5π12
Check.
x = pi/12 = 180/12 = 15^@x=π12=18012=15∘ --> 2x = 30^@2x=30∘ --> 4x = 60^@4x=60∘
sin 30 + sin 60 = 1/2 + sqrt3/2sin30+sin60=12+√32
cos 30 + cos 60 = sqrt3/2 + 1/2cos30+cos60=√32+12. Proved
