Apply the trig identity:
sin a + sin b = 2sin ((a +b)/2)cos ((a - b)/2)sina+sinb=2sin(a+b2)cos(a−b2)
sin 2x + sin 4x = 2sin ((6x)/2)cos ((2x)/2) = 2sin 3x.cos x sin2x+sin4x=2sin(6x2)cos(2x2)=2sin3x.cosx
(sin 2x + sin 4x) + sin 3x = sin 3x(2cos x + 1) = 0(sin2x+sin4x)+sin3x=sin3x(2cosx+1)=0
Either one of the 2 factors must be zero.
a. sin 3x = 0 .
Trig table give 3 solution arcs:
3x = 0 --> x = 0
3x = pi3x=π --> x = pi/3x=π3
3x = 2pi3x=2π --> x = (2pi)/3x=2π3
b. 2cos x + 1 = 0 --> cos x = - 1/2cosx=−12
Trig table and unit circle give 2 solution arcs:
x = 2pi/3x=2π3 and x = - (2pi)/3x=−2π3 --> or x = (4pi)/3x=4π3 (co-terminal)
Answers for (0, 2pi)(0,2π):
0, pi/3, (2pi)/3, (4pi/3)0,π3,2π3,(4π3)
