MATH THAT I NEED HELP WITH

Question

Solve  for the two smallest positive solutions A and B, with A < B

Answer 1

sin4x=4/5=.8

4x₁=sin^-1.8 +2πk, or x₁=(sin^-1.8)/4 + 2πk/4=(sin^-1.8)/4 + πk/2 k=0,1,-1,2,-2,...

4x₂= π - sin^-1.8 +2πk or x₂=(2k+1)*π/4-(sin^-1.8)/4 k=0,1,-1,2,-2,...

for sinB=j, that is procedure, B₁=(sin^-1j)+2πk and B₂=π-sin^-1j+2πk=(2k+1)π-sin^-1j k=0,1,-1,2,-2,...

now for what is two smallest angles where sin4x=.8

smallest is x₁ with k=0 or x=(sin^-1.8)/4

and with k=0 for x₂ x=π/4-(sin^-1.8)/4 , which I can prove is less than x₁ with k=1 or x=(π/2)+(sin^-1.8)/4

smallest angles are x=(sin^-1.8)/4, and x=π/4-(sin^-1.8)/4

Answer 2

5 sin 4x = 4

sin 4x = 0.8

4x ≈ 53.13

4x ≈ 126.87