Question #54096

1 Answer
Oct 9, 2015

x = npi/2color(white)("XXXXXXX") or
x = (2pi)/3+n2picolor(white)("XXX") or
x=(4pi)/3+n2pi

Explanation:

Since sin (3x) + sin(x)
= 2 sin (2x) cos(x) color(white)("XXX")(See note below if this is not obvious)

=> sin(x) + sin(2x) + sin(3x)
color(white)("XXX")= [sin(3x)+sin(x)] + sin(2x)
color(white)("XXX")=[2sin(2x)cos(x)] +sin(2x)
color(white)("XXX") = sin(2x)( 2cos(x) + 1)

=> the equation reduces to
sin( 2x) = 0 or cos( x) = - 1/2

If sin( 2x) = 0
color(white)("XXX")=> 2x = npi .

If cos( x) = - 1/2
color(white)("XXX")(within the interval [0,360^@) = [0,2pi))
color(white)("XXX")=> x = 120^@ or x = 240^@
color(white)("XXXXXX") = ((2pi)/3) or ((4pi)/3)

Note
The transformation sin(3x)+sin(x) = 2sin(2x)cos(x)
is based on the formula:
sin(A)+sin(B) = 2sin((A+B)/2)*cos((A-B)/2)