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Is a=2a=2 the only solution to the equation below? aC:xRcosax=acosax1\boxed{a\in\mathbb{C}:\forall x\in\mathbb{R} \cos ax=a\cos^ax-1}

#Geometry

Note by Zakir Husain
3 months, 1 week ago

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Comments

Yes, a=2a = 2 is only solution to above equation.

As above equation is valid for all xRx \in R, so it should also be valid for x=0x = 0. Putting x=0x = 0

cosa(0)=acosa01\cos a(0) = a\cos^a 0 - 1

1=a11 = a - 1

a=2a = 2

As only one value of aa comes out in solution, only one exists. Hope it helps. :)

@Zakir Husain

Aryan Sanghi - 3 months, 1 week ago

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Clever solution. :)

Zakir Husain - 3 months, 1 week ago

But if a(0,1)a\in (0,1) then multiple value of 1a1^a are there (not considering principle values)

Zakir Husain - 3 months, 1 week ago

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Let a(0,1)a \in (0,1), then

1=a.1a11 = a.1^a - 1 2=a.1a2 = a.1^a

1a=2a1^a = \frac2a

As a(0,1)a \in (0,1), RHS is real, positive and >2\gt 2. But, LHS will be negative or 11 or non real. So, a(0,1)a \in (0,1) won't satisfy. Similar proof is also valid for a(1,0)a \in (-1,0). Hope it helps. :)

Aryan Sanghi - 3 months, 1 week ago

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@Aryan Sanghi But if aCR  1aa\in\mathbb{C}-\mathbb{R}\space\space 1^a will have \infty values.

For example : (1)i=(ein2π)i=en2π,nZ(1)^i=(e^{in2\pi})^i=e^{-n2\pi},n\in\mathbb{Z} (Not considering principle values)

Zakir Husain - 3 months, 1 week ago

Very cool. The double angle formula cannot be generalized. :)

David Stiff - 3 months, 1 week ago
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