How do you solve sin(2x) = —√3/2 and find all the solutions in the interval [0,2π)?

Answer 1

If sin (Θ) = A, then A = arcsin (A). You would get this from your calculator.

The interval 0 to 2π describes a circle. Plot any arbitrary angle and then note that if you plot another angle with the same y value, but with a different sign for x, you have a mirror image around the y-axis. The 2 angles will have the same sine, but the cosines will be the negative of each other. Similar logic for a pair of angles mirrored about the x-axis, except now the cosines are the same.

So---as you go around a circle, there are exactly 2 places where the sine of the angle is the same, and exactly 2 (other) places where the cosines are the same

Answer 2

pi is negative in the third and fourth quadrants

look on the unit circle

sin=-√3/2 at 4∏/3 or 240°

also, sin=-√3/2 at 5∏/3 or 300°

therefore 2x=240° and x=120°

also 2x=300° and x=150°