If x2-2xcosθ+1=0 , then the value of x2n-2xncosnθ+1, n∈N, is equal to -
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a
cos2nθ
b
sin2nθ
c
0
d
some real number greater than 0
answer is C.
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Detailed Solution
x2-2xcosθ+1=0∴x=2cosθ±4cos2θ-42=cosθ±isinθ Let x=cosθ+isinθ∴ x2n-2xncosnθ+1=cos2nθ+isin2nθ-2(cosnθ+isinnθ)cosnθ+1=cos2nθ+1-2cos2nθ+i(sin2nθ-2sinnθcosnθ)=0+i0=0