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How do you solve arcsin(sqrt(2x))=arccos(sqrtx)? | HIX Tutor

We have to take the sine or the cosine of both sides. Pro Tip: choose cosine. It probably doesn't matter here, but it's a good rule.
So we'll be faced with <mathjax># cos arcsin s #</mathjax>
That's the cosine of an angle whose sine is <mathjax>#s#</mathjax>, so must be
<mathjax># cos arcsin s = pm \sqrt{1 - s^2} #</mathjax>
Now let's do the problem
<mathjax># arcsin (sqrt{2x }) = arccos(\sqrt x)#</mathjax>
<mathjax>#cos arcsin (\sqrt{2 x}) = cos arccos ( \sqrt{x})#</mathjax>
<mathjax>#\pm \sqrt{1 - (sqrt{2 x})^2 } = sqrt{x}#</mathjax> 
We have a <mathjax>#pm#</mathjax> so we don't introduce extraneous solutions when we square both sides.
<mathjax># 1 - 2 x = x #</mathjax>
<mathjax># 1 = 3x #</mathjax>
<mathjax>#x = 1/3 #</mathjax>
Check:
<mathjax># arcsin \sqrt{2/3} stackrel?= arccos sqrt{1/3}#</mathjax>
Let's take sines this time.
<mathjax>#sin arccos sqrt{1/3} = pm sqrt{1 - (sqrt{1/3})^2} =pm sqrt{2/3}#</mathjax>
Clearly the positive principal value of the arccos leads to a positive sine.
<mathjax># = sin arcsin sqrt{2/3) quad sqrt#</mathjax>

We have to take the sine or the cosine of both sides. Pro Tip: choose cosine. It probably doesn't matter here, but it's a good rule.
So we'll be faced with <mathjax># cos arcsin s #
That's the cosine of an angle whose sine is <mathjax>#s#</mathjax>, so must be
<mathjax># cos arcsin s = pm \sqrt{1 - s^2} #
Now let's do the problem
<mathjax># arcsin (sqrt{2x }) = arccos(\sqrt x)#
<mathjax>#cos arcsin (\sqrt{2 x}) = cos arccos ( \sqrt{x})#
<mathjax>#\pm \sqrt{1 - (sqrt{2 x})^2 } = sqrt{x}#
We have a <mathjax>#pm#</mathjax> so we don't introduce extraneous solutions when we square both sides.
<mathjax># 1 - 2 x = x #
<mathjax># 1 = 3x #
<mathjax>#x = 1/3 #
Check:
<mathjax># arcsin \sqrt{2/3} stackrel?= arccos sqrt{1/3}#
Let's take sines this time.
<mathjax>#sin arccos sqrt{1/3} = pm sqrt{1 - (sqrt{1/3})^2} =pm sqrt{2/3}#
Clearly the positive principal value of the arccos leads to a positive sine.
<mathjax># = sin arcsin sqrt{2/3) quad sqrt#

Madelyn Adams

To solve the equation arcsin(sqrt(2x)) = arccos(sqrt(x)), first, express both sides in terms of the same trigonometric function, then apply trigonometric identities and solve for x. 

arcsin(sqrt(2x)) = arccos(sqrt(x))
sin(arcsin(sqrt(2x))) = sin(arccos(sqrt(x)))
sqrt(2x) = cos(arccos(sqrt(x)))
sqrt(2x) = sqrt(x)

Square both sides of the equation to eliminate the square roots:

2x = x

Now, solve for x:

2x - x = 0
x = 0

However, you should check whether x = 0 is a valid solution by substituting it back into the original equation, as it might not satisfy the domain of the inverse trigonometric functions. 

arcsin(sqrt(2*0)) = arcsin(0) = 0
arccos(sqrt(0)) = arccos(0) = π/2

Since arcsin(0) = 0 and arccos(0) = π/2, the solution x = 0 is valid. Therefore, the solution to the equation arcsin(sqrt(2x)) = arccos(sqrt(x)) is x = 0.

How do you solve #arcsin(sqrt(2x))=arccos(sqrtx)#?