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

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Answer 1 · 2018-05-14T08:51:44

#x = 1/3 #

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 # cos arcsin s #

That's the cosine of an angle whose sine is #s#, so must be

# cos arcsin s = pm \sqrt{1 - s^2} #

Now let's do the problem

# arcsin (sqrt{2x }) = arccos(\sqrt x)#

#cos arcsin (\sqrt{2 x}) = cos arccos ( \sqrt{x})#

#\pm \sqrt{1 - (sqrt{2 x})^2 } = sqrt{x}#

We have a #pm# so we don't introduce extraneous solutions when we square both sides.

# 1 - 2 x = x #

# 1 = 3x #

#x = 1/3 #

Check:

# arcsin \sqrt{2/3} stackrel?= arccos sqrt{1/3}#

Let's take sines this time.

#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.

# = sin arcsin sqrt{2/3) quad sqrt#