# Trig Equation

TD
Homework Helper
cscott said:
Wait! Let me just check my understanding: the solution to the equation I posted can be either the one including pi/3 or the one inlcluding 4pi/3 and each solution has to encorporate the negative and positive sqrt(3)/2?

I say this because pi/3 + pi gives -sqrt(3)/2 but I got the pi/3 in the first place using sin x = +sqrt(3)/2

Is this correct? It doesn't seem like it...
From the positive root, the solution pi/3 and its complementary solution pi-pi/3 = 2pi/3 follow.
From the negative root, the solution -pi/3 (or 5pi/3) and its complementary solution 4pi/3 follow.

You see?

TD said:
From the positive root, the solution pi/3 and its complementary solution pi-pi/3 = 2pi/3 follow.
From the negative root, the solution -pi/3 (or 5pi/3) and its complementary solution 4pi/3 follow.

You see?

I understand that there is two solutions for each root but I don't see how that translates into the one solution per root. How do you get 2pi/3 from pi/3 + pi * k? Shouldn't I be able to put any integral value for "k" and get a positive sqrt(3)/2 answer in that case?

Last edited:
TD
Homework Helper
For the positive root, we had: pi/3 and 2pi/3
For the negative root, we had: 4pi/3 and 5pi/3

Of course, all +2k*pi.

But now notice that 4pi/3 = pi/3 + pi and 5pi/3 = 2pi/3 + pi.
So you need to 'cross' the solutions of both roots, take them all together.
Then you can leave out the two redundant ones and use +k*pi instead of +2k*pi.

I get it! I get it! Thanks again for being patient with me.

It's like this was a one on one (fast posts) :tongue:

TD
Homework Helper