Math problems of trigonometric function. Please help me!?

Question

ƒ(x) = A sin (π/2 x) + B ; where A=2 ; B=1
solve the equation ƒ(x) = 2 for 0 ≤ x ≤ 5

Answer 1

Just make the appropriate substitutions:

2 = 2 sin (π/2 x) + 1

And solve:

1 = 2 sin (π/2 x)
sin (π/2 x) = 1/2
π/2 x ∈ {π/6 + 2πk: k∈ℤ}∪{5π/6 + 2πk: k∈ℤ}
x∈{1/3 + 4k: k∈ℤ}∪{5/3 + 4k: k∈ℤ}

Now just find the values of x which lie between 0 and 5:

x∈{1/3, 13/3, 5/3}

And we are done.

Answer 2

ƒ(x) = 2 sin (π/2 x) + 1
there are a lot of ups and down of the curve from 0 to 1
. . . (4 ups . .. and . . 4 downs )
x . . . . . . . . y
0.20 . . . .. 3 . .. one of the high
0.2727 . . . .0 . . . intersection with . x-axis
1/3 . . . . .. . -1 . . . lowest point
0.4286 . . . . 0 . .. intersection with x-axis
0.5. . . . . . .0.963
0.60 . . . . . 2
0.70 . . . . . 2.555
0.80 . . . . . 2.841
0.90 . . . . . 2.969
1 . . . . . . . . 3 . . . . . . highest point
1.2 . . . . . . 2.93
1.4 . . . . . . 2.80
1.8 . . . . . 2.533
2 . . . . . . . 2.412
3 . . . . . . . 2
4 . . . . . . . 1.767
5 . . . . . . . 1.615
10 . . . . . . . 1.312
20 . . . . . . . 1.157

Answer 3

F(x) = 2

2 = (2)sin((pi)/2x) + 1


1 = (2)sin((pi)/2x)

(1/2) = sin((pi)/2x)

arcsin(1/2) = pi/2x

**** arcsin(1/2) = (pi)/4

(pi)/4 = (pi)/2x

====> 4 = 2x

x = 2.