Question

solve for x --- arcsin(2x)^1/2 = arccos(x)^1/2

Answer 1

Nobody?

Answer 2

im getting x is:
\[\frac{1}{\sqrt{5}}\], but i need to verify my answer. one sec.

Answer 3

http://www.wolframalpha.com/input/?i=solve+arcsin(2x)^1%2F2+%3D+arccos(x)^1%2F2
This is the solution. I'll let joe help you do the analytic part.

Answer 4

You're right joe^^^^

Answer 5

book says x=1/3......

Answer 6

Then you have type the problem wrong :P

Answer 7

is it this?
\[(\sin^{-1}(2x))^{\frac{1}{2}} = (\cos^{-1}(x))^{\frac{1}{2}}\]

Answer 8

cause thats what i solved.

Answer 9

arcsin and sin-1 are the same yes?

Answer 10

right.

Answer 11

note, what i typed is different than:
\[\sin^{-1}((2x)^\frac{1}{2}) = \cos^{-1}((x)^\frac{1}{2})\]

Answer 12

arcsin\[\sqrt{2x}\]+arcos\[\sqrt{x}\] solve for x

Answer 13

ah, its the second one then.

Answer 14

I don't know why it's put in a column like that...

Answer 15

It automatically spaces it that way.

Answer 16

its like that, but i get the problem now.

Answer 17

so the way i solved it was i said let:\[y = \sin^{-1}(\sqrt{2x}), y = \cos^{-1}(\sqrt{x})\] we can set them both equal to y since they both equal each other. Taking the sine of both sides of the first equation, and cosine of both sides in the second gives:
\[\sin(y) = \sqrt{2x}, \cos(y) = \sqrt{x} \iff \sin^2(y) = 2x, \cos^2(y) = x\]Now using the trig identity:\[\sin^2(y)+\cos^2(y) = 1\]we obtain:
\[2x+x = 1 \iff 3x = 1 \iff x = \frac{1}{3}\]

Answer 18

steps are unavailable on the wolfram website.....

Answer 19

very good. thank you. it makes prefect sense now