How do you make y= sin(sqrt(5x + 3)) in terms of y? :/

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1. Yes, I don't know this. =.=

So how to you move sine to the y side?

You can't say: arcsin(y) = sqrt(5x + 3), can you...?

I don't know how to move the sin!
Help!
Thank you! :)
 
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Lo.Lee.Ta. said:
1. Yes, I don't know this. =.=

So how to you move sine to the y side?

You can't say: arcsin(y) = sqrt(5x + 3), can you...?

I don't know how to move the sin!
Help!
Thank you! :)

You can do that as long as you realize it will only be valid for -pi/2<=sqrt(5x+3)<=pi/2.
 
Hi Lo.Lee.Ta.! :smile:

Generally, the equation
$$y=sin(\theta)$$
is equivalent to
$$\theta=\arcsin(y)+ 2k\pi \quad \vee \quad \theta=\pi - \arcsin(y) + 2k\pi$$
where ##k## is an integer and ##\theta## is an angle in radians.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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