find ALL solutions for the next two problmes.?

Question

1. sin^2(3x)-1=0

2. 8cos^3x-1=0

Answer 1

1. sin^2(3x)-1=0
sin^2(3x) = 1
sin3x = +/- 1
3x = 90 or 270
x = 30 or 90 degrees

2. 8cos^3x-1=0
8cos^3x =1
2cos x = 1
cos x= 1/2
x = 60 or 300 degrees

Answer 2

1. Rewrite: sin^2(3x) = 1 hence sin(3x) =+-1

When 3x = pi/2, sin(3x) = 1. Hence one solution is x = pi/6. Observe that the period of sin(3x) is 2pi/3. So all solutions are of the form pi/6 + n2pi/3 where n is an integer.
When 3x = 3pi/2, sin (3x) = -1. Hence one solution is x = pi/2. It follows that sin(3x) = -1 whenever x = pi/2 + n2pi/3 where n is an integer.
Combining these results together, we get that sin^2(3x) = 1 whenever x = pi/6 + npi/3 where n is an integer (i.e. n = 0, 1, -1, 2, -2...)

2. I assume you mean 8cos^3(x) - 1 = 0. Then cos^3(x) = 1/8 or cos(x) = 1/2. This occurs whenever x = pi/3 or x = -pi/3 = 2pi - pi/3 = 5pi/3. Generalizing the two solutions, we get (1) x = pi/3 + n2pi and (2) x = 5pi/3 + n2pi where n is an integer.