Solving Trig Equations

[reverdybrune]

solve these for x:
sin(x/2)=-1/2
2cos(x/3)+1=0

Note: re-posted from my old OO.net blog

Comments

[reverdybrune]

tommyedison (on OO.net) commented:

First one is -60 degrees
Second one is 360 degrees

[reverdybrune]

Actually, there are an infinite number of solutions in this context. The first can be written as 7pi/3+ 4k pi and 11pi/3 + 4k pi where k is an integer.

The second can be written as 2pi + 6k pi and 4pi +6k pi where k is an integer.

tommyedison's answers are correct, however they are not the complete answers.

If the question was sin^-1(x/2)=-1/2 then -60 degrees or pi/3 would be the only answer. However, when the equation is a normal sine function (not inverse) there is no restriction on the domain.

[reverdybrune]

realitycheck44 (on OO.net) commented:

QUOTE(non-contradictor @ May 10 2005, 06:39 PM):
If the question was sin^-1(x/2)=-1/2 then -60 degrees or pi/3 would be the only answer. However, when the equation is a normal sine function (not inverse) there is no restriction on the domain.

Ha, I missed this earlier. Anyway, I got the same answers as tommy. (I didn't post because there's really no point once the question is solved.) When this question is asked, most textbooks limit the domain to -2pi to +2pi. That's why his, and my, answers were not continuous answers.

Anyway, tommy, is your textbook the same as mine?

Zak

[reverdybrune]

That's quite interesting. At IMSA, the teachers have created their own math curriculum, and we don't use textbooks. I hope I don't run into too many problems because they teach differently...