How do you solve for the exact solutions in the interval [0,2pi] of #sin 2x=sinx#?
1 Answer
Mar 4, 2016
Explanation:
Using appropriate
#color(blue)" Double angle formula " # • sin2A = 2sinAcosA
hence : 2sinxcosx = sinx
and 2sinxcosx - sinx = 0
Take out common factor of sinx
thus : sinx(2cosx - 1 ) = 0
and so sinx = 0 or cosx =
#1/2 # sinx = 0
#rArr x = 0 or 2pi # cosx
# = 1/2 rArr x = pi/3 or (2pi - pi/3 ) = (5pi)/3 # These are the solutions in the interval [0 ,
#2pi] #
