Math Help - Trig problem

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

Originally Posted by silly_girl

Find all soluctions of the equations:
(a) SinX=-1

As I noted in another thread, Sin(x) and sin(x) are technically two different functions. The common trig functions are written without capital letters.

sin(x) = -1

We know that, for 0 <= x < 2(pi), that sin(x) = -1 for only x = 3(pi)/2. This solution is periodic with a period of 2(pi), so the solution will be:
x = 3(pi)/2 + 2(pi)*n, where n is any integer.

-Dan

Originally Posted by silly_girl

(b) Cos u + Sin 2u = 0

Note that sin(2u) = 2*sin(u)*cos(u) so:
cos(u) + 2*sin(u)*cos(u) = 0 <-- Factor the common cos(u):

cos(u)*[1 + 2*sin(u)] = 0

Now either
cos(u) = 0 ==> u = (pi)/2, 3(pi)/2

or

1 + 2*sin(u) = 0 ==> sin(u) = -1/2 ==> u = 7(pi)/6, 11(pi)/6

Again, each of these solutions are periodic, so the final solution is:
u = (pi)/2 + 2(pi)*n, 7(pi)/6 + 2(pi)*n, 3(pi)/2 + 2(pi)*n, 11(pi)/6 + 2(pi)*n, where n is any integer.

-Dan

Originally Posted by silly_girl

(c) 2Sint Cost -Sint = 0

Similar to b):
sin(t)*[2*cos(t) - 1] = 0

sin(t) = 0 ==> t = 0, (pi)

2*cos(t) - 1 = 0 ==> cos(t) = 1/2 ==> t = (pi)/3, 5(pi)/3

So for 0 <= t < 2(pi)

t = 0, (pi)/3, (pi), 5(pi)/3
(You can add the 2(pi)*n's this time.)

-Dan

Originally Posted by silly_girl

(d) (0, 360)

What's this question again?

-Dan