Question

pls explain for me why (3/2pi)=sin2x => x= (1/2)arcsin(3/2pi)

Answer 1

\[\frac{ 3 }{ 2\pi }=\sin2x => x=\frac{ 1 }{ 2 }\arcsin(\frac{ 3 }{ 2\pi })\]

Answer 2

cos(3y) dy = -sin(2x) dx

Integrating both sides, we have

INT cos(3y) dy = INT -sin(2x) dx

(1/3)sin(3y) = (1/2)cos(2x) + C

Given the initial condition y(pi/2) = pi/3, solve for C.

(1/3)sin(3*pi/3) = (1/2)cos(2*pi/2) + C

0 = (1/2)*(-1) + C

C = 1/2

Answer 3

@dangbeau there?

Answer 4

yes

Answer 5

okay\[\frac{ 3 }{ 2\pi } =\sin(2x)\]

taking sine inverse we get
\[\sin^{-1}\frac{ 3 }{ 2\pi } = 2x\]
we want x right so subjecting it we get\[x = \frac{ 1 }{2}\sin^{-1} (\frac{ 3 }{ 2\pi })\]

Answer 6

IS it clear to u now ?

Answer 7

i got it, thank you ^^