It's problem 20 on this website:
http://www.indianamath.org/includes/down…
http://www.indianamath.org/includes/down…
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x+1/x = 2cosθ
x^2 +1 = 2xcosθ ==> x^2 - (2cosθ)x +1 = 0
x = (2cosθ +/- √4(cosθ)^2 - 4)/2
= (2cosθ +/- √4[(cosθ)^2 -1])/2
= (2cosθ +/- √-4(sinθ)^2)/2
= (2cosθ +/- i2sinθ)/2
= cosθ +/- isinθ
x^n = cosnθ +/- i sinnθ
1/x^n = x^(-n) = cos(-nθ) +/-isin(-nθ) = cosnθ -/+ isinnθ
(note that it is +/- for x^n and -/+ for 1/x^n)
x^n + 1/x^n = 2cosnθ
x^2 +1 = 2xcosθ ==> x^2 - (2cosθ)x +1 = 0
x = (2cosθ +/- √4(cosθ)^2 - 4)/2
= (2cosθ +/- √4[(cosθ)^2 -1])/2
= (2cosθ +/- √-4(sinθ)^2)/2
= (2cosθ +/- i2sinθ)/2
= cosθ +/- isinθ
x^n = cosnθ +/- i sinnθ
1/x^n = x^(-n) = cos(-nθ) +/-isin(-nθ) = cosnθ -/+ isinnθ
(note that it is +/- for x^n and -/+ for 1/x^n)
x^n + 1/x^n = 2cosnθ
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Formally the correct answer is B, although I do not immediately see how to prove it.
However, it is a bit of a bad question because the left side x + 1/x is always bigger than or equal to 2 in absolute value, whereas the right side, 2cos(theta) is always less than or equal to 2. So the two sides can only be equal for x=1 when both sides are 2 and all the answers are correct. Otherwise, the only possible values of x are complex numbers.
Nonetheless, formally at least B is correct.
However, it is a bit of a bad question because the left side x + 1/x is always bigger than or equal to 2 in absolute value, whereas the right side, 2cos(theta) is always less than or equal to 2. So the two sides can only be equal for x=1 when both sides are 2 and all the answers are correct. Otherwise, the only possible values of x are complex numbers.
Nonetheless, formally at least B is correct.