Question #2898e

1 Answer
Aug 30, 2016

#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^@#