How do you solve #sin2x+sin4x= 0#?

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Answer 1 · 2015-06-12T17:58:27

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

Use the trig identity: sin a + sin b = #2sin ((a + b)/2) cos ((a -b)/2)#

f(x) = 2sin 3x.cos x = 0

Next solve sin 3x = 0 and solve cos x = 0.

a. sin 3x = 0 --> 3x = 0 and 3x = pi - 0 = pi --> #x = pi/3#

and 3x = 2pi --># x = 2pi/3#

b. cos x = 0 --> #x = pi/2# and #x = 3pi/2#

Answers within interval #(0, 2pi): #
#0, pi/3, pi/2, (2pi)/3, (3pi)/2 and 2pi#