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???

To find all the solutions of the equation sin 2x = -sqrt(3)/2 in the interval (0, 2pi), you correctly identified the angles 4pi/3 and 5pi/3 as solutions.

Next, considering the given equation sin 2x = -sqrt(3)/2 and knowing that the sine function has a period of 2pi, you need to account for all possible values of x within the interval (0, 2pi).

To do this, you correctly added the 2npi term to account for all possible solutions due to the periodic nature of the sine function.

So far, you have 2x = 4pi/3 + 2npi and 2x = 5pi/3 + 2npi.

Now, to solve for x, you divide both sides of the equations by 2:

x = (4pi/3 + 2npi)/2 and x = (5pi/3 + 2npi)/2.

Simplifying these expressions, you get:

x = 2pi/3 + npi and x = 5pi/6 + npi.

Therefore, the correct solutions within the interval (0, 2pi) are:

x = 2pi/3 + npi and x = 5pi/6 + npi, where n is an integer.

So, you have correctly identified the solutions and expressed them in terms of x within the specified interval. Well done!