Don't be Fooled 3

Geometry Level 2

Find the number of real solutions to the equation $\large e^{\ln(x)}=\cos(\sin^{-1}(x)).$

3 0 2 1

1 solution

Hjalmar Orellana Soto
Aug 18, 2015

$e^{ln(x)}=\cos(\sin^{-1}(x))$ $=>x=\sqrt{1-x^{2}}$ $=>x^{2}=1-x^{2}$ $=>x^{2}=\frac{1}{2}$ Because of the logarithm at the begining we can't take the negative solution to this ecuation, so, there's only one real solution.

how can $\cos(\sin^{-1}(x))$ became $\sqrt{1-x^2}$ ?

Alvin Willio - 5 years, 9 months ago

If you draw a right triangle with hypotenuse $1$ and foots $x$ and $\sqrt{1-x^{2}}$ you'll find that there is an angle $\theta$ such that $\sin(\theta) = x$ , then $\theta = \sin^{-1} (x)$ . In that triangle we have $\cos(\theta)=\sqrt{1-x^{2}}$ , but $\theta=(\sin^{-1}(x))$ , so $\cos(\sin^{-1}(x))=\sqrt{1-x^{2}}$ . Is an usefull property.

Hjalmar Orellana Soto - 5 years, 9 months ago

Oh, I see.. Thank you..

Alvin Willio - 5 years, 9 months ago

Don't be Fooled 3

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