sin5x + sinx = 0?

Question

how do you solve??

Answer 1

sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
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sin(A+B) + sin(A-B) = 2sinAcosB

sin(5x) + sin(x) = 0
sin(3x+2x) + sin(3x-2x) = 0
2sin(3x)cos(2x) = 0
sin(3x)cos(2x) = 0

sin(3x) = 0
3x = nπ
x = nπ/3 where n is an integer

OR

cos(2x) = 0
2x = (2n+1)π/2
x = (2n+1)π/4 where n is an integer

Answer 2

Sin5x Sinx

Answer 3

sin x = 0
x = 0, pi, 2 pi
and scale any of values by 5 x still = 0
for example: sin5pi=0, sin 10pi
so sin x = 0 and sin 5x = 0 -> x = 0, pi, 2 pi
since 0+0 = 0 -> sin5x + sinx = 0
also, those values for x are only 0<=x<=2pi, but the sine curve is infinite so there are infinite vales for what x can be

Answer 4

sin A + sin B
2 sin [ (A + B) / 2 ] cos [ (A - B) / 2 ]
sin 5x + sin x = 0
2 sin 3x cos 2x = 0
sin 3x = 0 , cos 2x = 0

3x = k π for k = 0 , 1 , 2 , 3 , -------
x = k π/3

2x = k π/2 , k 3π/2
x = (k/2)π/2 , (k/2) 3π/2