Solving cos(θ)^x+sin(θ)^x=1 for x: How many real solutions exist?

In summary, the conversation discusses how to prove that 2 is the only possible solution for the equation cos(θ)^x + sin(θ)^x = 1 for all values of θ. It is suggested to choose a particular value of θ, such as one where cosθ = sinθ, and prove that 2 is the only solution. This would prove the equation for all values of θ. The conversation ends with the speaker expressing their gratitude and stating that they will try to complete the proof on their own.
  • #1
vijayramakrishnan
90
0

Homework Statement


cos(θ)^x+sin(θ)^x=1
find the number of real values of x satisfying this equation

Homework Equations


none

The Attempt at a Solution


the answer given is 1 which is 2.which is obvious but i don't know how to prove that it is the only solution available
 
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  • #2
vijayramakrishnan said:
but i don't know how to prove that it is the only solution available
You want to prove there is no x, other than 2, which gives equality for all values of θ.

Choose any particular θ and show that the only x-value for that particular θ which gives equality is x=2 (that would prove it, right?)

What particular value of θ might it be easy to show this for?
 
  • #3
Nathanael said:
You want to prove there is no x, other than 2, which gives equality for all values of θ.

Choose any particular θ and show that the only x-value for that particular θ which gives equality is x=2 (that would prove it, right?)

What particular value of θ might it be easy to show this for?
thank you for replying sir,i want to prove that 2 is the only solution possible.
 
  • #4
vijayramakrishnan said:
thank you for replying sir,i want to prove that 2 is the only solution possible.
When you say "the only solution possible," what you really mean is, "the only solution possible for all values of θ" right?

If you were to choose a particular value of θ and prove that 2 is the only solution for that θ, then that would prove what you want, because if x is not 2 then x does not solve it for that particular θ you chose.
This means you only have to prove it for any single value of θ. Do you agree?

I didn't want to hint at it too much, but if you choose the θ such that cosθ=sinθ then you can factor together the exponent x and complete the proof.
 
  • #5
Nathanael said:
When you say "the only solution possible," what you really mean is, "the only solution possible for all values of θ" right?

If you were to choose a particular value of θ and prove that 2 is the only solution for that θ, then that would prove what you want, because if x is not 2 then x does not solve it for that particular θ you chose.
This means you only have to prove it for any single value of θ. Do you agree?

I didn't want to hint at it too much, but if you choose the θ such that cosθ=sinθ then you can factor together the exponent x and complete the proof.

thank you sir, from there i will try on my own
 

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