Using the identity #sin(2x) = 2sin(x)cos(x)#, we have
#sin(x)cos(x) = 1/3#
#=> 2sin(x)cos(x) = 2/3#
#=> sin(2x) = 2/3#
Note that we must also consider the identity #sin(x) = sin(pi-x)# in order to get all solutions. That gives
#sin(pi-2x) = 2/3#
Applying the inverse sine function to both sides, we get
#2x = arcsin(2/3) + 2npi#
or
#pi-2x = arcsin(2/3) + 2npi#
We include the #2npi# as sine is periodic with a period of #2pi#, meaning we can add or subtract integer multiples of #2pi# from the argument without changing the value of sine. The inverse sine function is defined to have the range #[-pi/2, pi/2]#, and thus will not provide all possible values.
Solving for #x#, we arrive at our solutions.
#x = arcsin(2/3)/2 + npi#
or
#x = -arcsin(2/3)/2+pi/2+npi#
Substituting in #n=0# and #n=1# into each solution set will give all of the provided answers. Note that substituting in any integer for #n# in either equation will give a solution.
