# How do you solve the initial-value problem #y'=sinx/siny# where #y(0)=π/4#?

You can separate yuor variables

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The solution is:

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To solve the initial-value problem ( y' = \frac{\sin(x)}{\sin(y)} ) with ( y(0) = \frac{\pi}{4} ), you can separate variables and then integrate.

First, separate variables by multiplying both sides by ( \sin(y) ) and dividing both sides by ( y' ): [ \sin(y) , dy = \sin(x) , dx ]

Next, integrate both sides: [ \int \sin(y) , dy = \int \sin(x) , dx ]

After integrating, you get: [ -\cos(y) = -\cos(x) + C ]

Where ( C ) is the constant of integration.

Now, apply the initial condition ( y(0) = \frac{\pi}{4} ): [ -\cos\left(\frac{\pi}{4}\right) = -\cos(0) + C ] [ -\frac{\sqrt{2}}{2} = -1 + C ] [ C = -\frac{\sqrt{2}}{2} + 1 ]

So, the particular solution to the initial-value problem is: [ -\cos(y) = -\cos(x) -\frac{\sqrt{2}}{2} + 1 ]

Solving for ( y ), you get: [ y = \arccos\left(\cos(x) +\frac{\sqrt{2}}{2} - 1\right) ]

This is the solution to the initial-value problem ( y' = \frac{\sin(x)}{\sin(y)} ) with ( y(0) = \frac{\pi}{4} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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