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Solve the equation.

(dy)/(dx)=(cos(x))/(sin(y))
Choose 1 answer:
(A)
y=arcsin(-sin(x))+C
(B)
y=arcsin(-sin(x)+C)
(C)
y=arccos(-sin(x))+C
(D)
y=arccos(-sin(x)+C)

Solve the equation. $\newline$ $\frac{d y}{d x}=\frac{\cos (x)}{\sin (y)}$ $\newline$ Choose $1$ answer: $\newline$ (A) $y=\arcsin (-\sin (x))+C$ $\newline$ (B) $y=\arcsin (-\sin (x)+C)$ $\newline$ (C) $y=\arccos (-\sin (x))+C$ $\newline$ (D) $y=\arccos (-\sin (x)+C)$

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Find derivatives of inverse trigonometric functions

Full solution

Q. Solve the equation.

\newline

\frac{d y}{d x}=\frac{\cos (x)}{\sin (y)}

\newline

Choose

1

answer:

\newline

(A)

y=\arcsin (-\sin (x))+C

\newline

(B)

y=\arcsin (-\sin (x)+C)

\newline

(C)

y=\arccos (-\sin (x))+C

\newline

(D)

y=\arccos (-\sin (x)+C)

Rewrite equation: We rewrite the equation as $\sin(y) \, dy = \cos(x) \, dx$ . Now we can integrate both sides: $\int \sin(y) \, dy = \int \cos(x) \, dx$ .
Integrate both sides: The integral of $\sin(y) \, dy$ is $-\cos(y)$ , and the integral of $\cos(x) \, dx$ is $\sin(x)$ . So we have $-\cos(y) = \sin(x) + C$ , where $C$ is the constant of integration. To solve for $y$ , we take the arccosine of both sides: $y = \arccos(-\sin(x) + C)$ . However, since the range of arccos is $[0, \pi]$ , and the range of arcsin is $[-\pi/2, \pi/2]$ , we should use arcsin to get the general solution. Therefore, we rewrite the equation as $-\cos(y)$ $0$ .

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