Solve for y: cos(3y) dy = -sin(2x) dx , y(pi/2) = pi/3?

Question

the solution manual says the answer is (pi - arcsin(3cos^2(x))) / 3. I got the arcsin(3cos^2(x)) / 3 part of it, but I don't understand the pi or the (-) sign. Please help?
also, it says the interval that the solution is defined is abs(x-pi/2) < 0.6155. can you explain this to me?

Tomato · Accepted Answer

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

So, we have (1/3)sin(3y) = (1/2)cos(2x) + (1/2). Now, recall the trigonometry identity cos(2x) = 2 cos^2(x) - 1 or cos^2(x) = (1/2)cos(2x) + (1/2). This gives us (1/3)sin(3y) = cos^2(x), or y = (1/3)arcsin[3cos^2(x)]

Now, the right hand side of the equation, cos^2(x) gives a +ve value, and we know that the sine function is positive in Quadrants 1 and 2. Thus y = (1/3)arcsin[3cos^2(x)] and also y = (1/3)[pi - arcsin[3cos^2(x)]]

Can't help with the 2nd part, sorry!

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Solve for y: cos(3y) dy = -sin(2x) dx , y(pi/2) = pi/3?

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