15 sin(2t) + 25 sin(t) = 0 Solve with the methods shown in this section exactly on the interval [0, 2π)

Answer 1

I am not sure of the method you are referring to but this is how I would solve

Use the identity sin(2x)= 2sin(x)Cos(x)

15(2sinxcosx) +25sinx=0

30sinxcosx +25sinx=0. Factor out 5sinx

5sinx(6cosx+5)=0 So

5sinx=0 Sinx=0 which would be at 0,pi and 2pi

And

6cosx+5=0. Cosx=(-5/6) take inverse cosine to get

x= 2.56 radians or -2.56radians

Answer 2

I don't know the method in the section, but here is the way I did it.

1. Substitute 2sin(t)cos(t) for sin(2t) to give
2. 15x2sin(t)cos(t) + 25sin(t) = 0 Factor out the sin(t) to get
3. sin(t) x (30cos(t) +25) = 0 As there are two factors, set each = 0
4. sin(t) = 0 and 30cos(t) + 25 = 0
5. The first equation gives t = 0 and 2π
6. The second equation gives cos(t) = -25/30 = -0.811
7. Using a calculator, and converting to radians gives approximately t = 0.814π and 1.628π.
8. You can put these t values into the original equation to confirm the results.