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???
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
Yes, your approach is correct up to a certain point.
To find all the solutions to the equation sin 2x = -√3/2 in the interval (0, 2π), you correctly determined the two principal solutions to be 4π/3 and 5π/3. Then, adding 2nπ to each principal solution gives you:
2x = 4π/3 + 2nπ
2x = 5π/3 + 2nπ
Now, to isolate x, you should divide both sides of each equation by 2:
x = (4π/3 + 2nπ) / 2
x = (5π/3 + 2nπ) / 2
Simplifying these expressions further gives you:
x = 2π/3 + nπ
x = 5π/6 + nπ
So, the correct solutions in the interval (0, 2π) are x = 2π/3 + nπ and x = 5π/6 + nπ, where n is an integer.