Question

Question #54096

Answer 1

`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)`