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.

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.(a) SinX=-1

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

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

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

factor out the sint(c) 2Sint Cost -Sint = 0

=> 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

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