Question

1) Find all the solutions of the equation in the interval (0,2pi).

sin 2x = -sqrt3/2

I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them.

2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi

Now I know that I have to divide by 2 but that is where I an not sure.

x = 4pi/6 + npi x=10pi/3 + npi

Is that correct???

Answer 1

Here is how I do these:
sin 2x = -√3/2
so by the CAST rule, 2x must be in quadrants III or IV
The angle in standard position (ignore the negative) is pi/3
so for III, 2x = pi + pi/3 = 4pi/3
for IV, 2x = 2pi - pi/3 + 5pi/3, you had both of those correct
Then x = 2pi/3 and 5pi/6

the period of sin 2x is pi, so adding/subtracting multiples pi to our two angles will yield more angles
since your domain is (0,2pi)
other solutions for x are 2pi/3 + pi = 5pi/3
and 5pi/6 + pi = 11pi/6

so x = 2pi/3, 5pi/3, 5pi/6 and 11pi/6
we would have