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.
