Math Help - Trig help needed

1. Trig help needed

Hi - I have been worrking on this problem for hours - and just am at a road block. Can anyone help me????? It would be so appreciated. This is my last problem and I just can't seem to get it started....

Find all soluctions of the equations: (a) SinX=-1 (b) Cos u + Sin 2u = 0 (c) 2Sint Cost -Sint = 0 (d) (0, 360)

any help with these would be greatly appreciated....

thanks
silly_girl

2. Originally Posted by silly_girl
Hi - I have been worrking on this problem for hours - and just am at a road block. Can anyone help me????? It would be so appreciated. This is my last problem and I just can't seem to get it started....

Find all soluctions of the equations: (a) SinX=-1 (b) Cos u + Sin 2u = 0 (c) 2Sint Cost -Sint = 0 (d) (0, 360)

any help with these would be greatly appreciated....

thanks
silly_girl
you're new, so i guess you didn't know, but double posting the same question is against the rules here. now you know, please don't do it again, it get's a lot of people upset.

(a) SinX=-1
look and the sine graph and notice that we have sinx = -1 when x = 3pi/2. but this is not the only value. the sine curve is periodic and therefore repeats values every period, which is every 2pi.

so all x's for which sinx = -1 is given by 3pi/2 + 2kpi, where k is a constant (do you undertsand why?)

Cos u + Sin 2u = 0
let's expand the sin2u using the double angle formula, we get

cosu + sinu*cosu + sinu*cosu = 0
=> cosu + 2sinu*cosu = 0
factor out the cosine
=> cosu(1 + 2sinu) = 0

since we now have 2 numbers being multiplied give zero, one or the other must be zero

=> cosu = 0 or 1 + 2sinu = 0

if cosu = 0
then u = pi/2 + kpi .........note that the cosine graph repeats zero twice every period, so the first positive value of u for which this happens is pi/2 and every pi after that (and before that)

if 1 + 2sinu = 0
=> sinu = -1/2

the first positive u for which this happens is 7pi/6 and then 11pi/6 and the corresponding values every period after that, look at the sine graph again

so u = 7pi/6 + 2kpi, u = 11pi/6 + 2kpi

so the solution to this problem is:
u = pi/2 + kpi, u = 7pi/6 + kpi, u = 11pi/6 + 2kpi

(c) 2Sint Cost -Sint = 0
factor out the sint
=> sint(2cost - 1) = 0 ..........now this is the same as the last question

=> sint = 0 or 2cost - 1 = 0

if sint = 0, then t = kpi

if 2cost - 1 = 0
=> cos t = 1/2
=> t = pi/3 + 2kpi, t = 5pi/3 + 2kpi

so the whole solution:

t = kpi, t = pi/3 + 2kpi, t = 5pi/3 + 2kpi

(d) (0, 360)
what are you talking about? there's no equation here

3. 1st - Thank you so much for helping me out!
2nd - OMG I am so sorry I double posted. I wasn't sure how to retract my post from the first forum.... so yes I double posted... VERY SORRY!!! I have been so frazzled over this stuff that I am just a mess at this point. I thought I was understanding things then when I came across this question I just went blank at where to even start! So thank you so much for simplifying things for me.
3rd - Forget (d) - that was from another question - see frazzled!!!!!!

Very appreciative of your assistance - have a terrific day!!!!!!

4. Originally Posted by silly_girl
1st - Thank you so much for helping me out!
2nd - OMG I am so sorry I double posted. I wasn't sure how to retract my post from the first forum.... so yes I double posted... VERY SORRY!!! I have been so frazzled over this stuff that I am just a mess at this point. I thought I was understanding things then when I came across this question I just went blank at where to even start! So thank you so much for simplifying things for me.
3rd - Forget (d) - that was from another question - see frazzled!!!!!!

Very appreciative of your assistance - have a terrific day!!!!!!

you too, good luck