We have: #sin(2 theta) = sin(theta)#; #(0, 360^(circ))#
Let's apply the double angle identity for #sin(theta)#:
#Rightarrow 2 sin(theta) cos(theta) = sin(theta)#
#Rightarrow 2 sin(theta) cos(theta) - sin(theta) = 0#
#Rightarrow sin(theta) (2 cos(theta) - 1) = 0#
Using the null factor law:
#Rightarrow sin(theta) = 0#
#Rightarrow theta = 0, 180^(circ) - 0#
#Rightarrow theta = 0, 180^(circ)#
#or#
#Rightarrow 2 cos(theta) - 1 = 0#
#Rightarrow cos(theta) = frac(1)(2)#
#Rightarrow theta = 60^(circ), 360^(circ) - 60^(circ)#
#Rightarrow theta = 60^(circ), 300^(circ)#
However, the interval given is #(0, 360^(circ))#.
So #0# will not be a solution to the equation.
Therefore, the solutions to the equation are #theta = 60^(circ)#, #theta = 180^(circ)# and #theta = 300^(circ)#.
