sin2 θ is the same as (sin θ)2. Factor, then solve. There are 3 solutions between 0 and 2π. One angle is on an axis, and the others are in Quadrants III and IV.
Answer
Example 4
Solve .
Hint
Wait until the very end to divide out the 2 from θ. Also, don't forget to find the equivalent reference angle in Quadrant IV.
Answer
for n = 0, ±1, ±2, etc.
for n = 0, ±1, ±2, etc.
Example 5
Solve 2sin (2θ) + 1 = 0 for .
Hint
If this were 2sin θ + 1 = 0, then the two angles between 0 and 2π that would satisfy this equation would be and . But that's not the equation we have to solve here. Too bad, so sad.
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for n = 0, ±1, ±2, etc.
for n = 0, ±1, ±2, etc.
.
and 
. But that's not the equation we have to solve here. Too bad, so sad.