Question

Question #2898e

Answer 1

`sin2x=sqrt3sinx`

`=>2sinxcosx-sqrt3sinx=0`

`=>sinx(2cosx-sqrt3)=0`

This gives `" "sinx=0 and cosx=sqrt3/2`

When `sinx=0`
the general suolution is

`x=npi," where "n in ZZ`

Since `" "0<=x<=360`

The solution within this limit can be had for `n =0 and 1`

So `" "x = 0 or pi`

Again when `cosx=sqrt3/2=cos(pi/6)`
the general suolution is

`x=2npi+-pi/6," where "n in ZZ`

Since `" "0<=x<=360`

The solution within this limit can be had for `n =0 and 1`

For n=0 `" "x = pi/6=30^@`

For n=1 `" "x = 2pi-pi/6=(11pi)/6=330^@`