Question

Solve the equation `sin2x+sin3x+sin5x=0` if `x` lies in the interval `0^@<x<180^@`?

Answer 1

`x=60^@,72^@,90^@,144^@`

Explanation:

`sin2x+sin3x+sin5x=0` can be written as

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

or `2sin((5x)/2)cos(x/2)+2sin((5x)/2)cos((5x)/2)=0`

i.e. `2sin((5x)/2)(cos(x/2)+cos((5x)/2))=0`

or `2sin((5x)/2)[2cos(((5x)/2+x/2)/2)cos(((5x)/2-x/2)/2)]=0`

or `4sin((5x)/2)cos((3x)/2)cosx=0`

i.e. either `sin((5x)/2)=0` i.e. `(5x)/2=npi` i.e. `x=(2npi)/5` i.e. `{72^@,144^@}` in the interval `0^@ < x < 180^@`

or `cos((3x)/2)=0` i.e. `(3x)/2=pi/2(2n+1)` i.e. `x=((2n+1)pi)/3` i.e. `{60^@}` in the interval `0^@ < x < 180^@`

or `cosx=0` i.e. `x=(2n+1)pi/2` i.e. `{90^@}` in the interval `0^@ < x < 180^@`

where `n` is an integer.

and in the interval `0^@ < x < 180^@`, `x=60^@,72^@,90^@,144^@`