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February 20

Trig functions

Hey I am in precalculus and need help (not answers) on solving two problems.

My job is to find the solutions of the equation that are in the interval [0, 2π) [couldn't find pi in the special characters.

sin 2t + sin t = 0 -- I know this can simplify to "2 sin t cos t + sin t = 0" but if I was trying to find solutions within 2π, where would I go from here? Am I allowed to factor sin t out?
cos u + cos 2u = 0 -- same problem. I know it simplifies to "cos u + cos^2 u = 0" or "cos u + 1 - 2 sin^2 u = 0", but same with the first problem, can I simplify cos u out?

How would I solve these problems? Thanks for your help!--Prowress (talk) 16:35, 20 February 2012 (UTC)[reply]

You can factor something, but not simply remove a common factor. So:

2 sin t cos t + sin t = 0 ⇒ sin t (2 cos t + 1) = 0 ⇒ sin t = 0 or 2 cos t + 1 = 0

The first equation of the last pair tells you t = nπ are solutions, n ∈ ℤ. The second equation of the pair gives futher solutions. Simply discarding a factor would hide half the solutions. — Quondum ^☏✎ 16:53, 20 February 2012 (UTC)[reply]

Double-check your first step in the second problem. --COVIZAPIBETEFOKY (talk) 17:13, 20 February 2012 (UTC)[reply]

So for #1, I got sin t = 0 and cos t = -1/2 but I cannot think of any configuration for both of them. --Prowress (talk) 23:08, 20 February 2012 (UTC)[reply]

Sorry, I meant by "configuration" that I cannot find any radians of pi that would fit both the answers.--Prowress (talk) 23:10, 20 February 2012 (UTC)[reply]

But you don't need a solution to both of them. You have two things which you're multiplying together, and they're supposed to make 0. So it's enough that one of them be 0.--130.195.2.100 (talk) 23:26, 20 February 2012 (UTC)[reply]

Meaning that for #1 you get sin t = 0 or cos t = -1/2 rather than sin t = 0 and cos t = -1/2 . Bo Jacoby (talk) 05:59, 21 February 2012 (UTC).[reply]