Question

Write `sin x +sin 3x` as a product. Hence,find all angles `x` in the interval`[0,2pi]` that satisfy `sin x +sin 3x - cos x =0`?

Answer 1

`x in{pi/12," "(5pi)/12," "pi/2," "(3pi)/2}`

Explanation:

First you should be aware that:
`sinA+sinB-=2sin((A+B)/2)cos((A-B)/2)`

So using the above result,
`sinx+sin3x=2sin((x+3x)/2)cos((x-3x)/2)`

`=2sin(2x)cos(-x)=color(blue)(2sin(2x)cos(x))`

Hence,

`sinx+sin3x-cosx=0`

Becomes,

`2sin(2x)cos(x)-cos(x)=0`

Factor out `cos(x)`

`=>cos(x)[2sin(2x)-1]=0`

Case 1 :
`cos(x)=0`

Thus , `x=+-pi/2+2npi" "` , `ninZZ`

Now we substitute all the values of `n` that will give us answers in in the interval `[0,2pi]`

`n=0" " =>x=color(orange)(pi/2)`
`n=1" " =>x=-pi/2+2pi=color(orange)((3pi)/2)`

Case 2:
`2sin(2x)-1=0`

`=>sin(2x)=1/2`

`=>2x=pi/6(-1)^n+npi" "` , `ninZZ`
`=>x=pi/12(-1)^n+npi/2`

Again, now we substitute all the values of `n` that give us the range we want,

`n=0" " =>x=color(orange)(pi/12)`

`n=1" " =>x=-pi/12+pi/2=color(orange)((5pi)/12)`