A) \[{{x}^{2}}+2x\,\cos \,\,n\theta +1=0\]
B) \[{{x}^{2}}-2x\,\cos \,\,n\theta +1=0\]
C) \[{{x}^{2}}-2x\,sin\,\,n\theta +1=0\]
D) \[{{x}^{2}}+2x\,sin\,\,n\theta +1=0\]
Correct Answer: B
Solution :
The roots of the given equation are \[x=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}=\cos \theta \pm i\,\,\sin \,\theta \] Let \[\alpha =cos\,\theta +i\,sin\theta \,\,\,\And \,\,\beta = cos \theta -i sin \theta \] Then \[{{\alpha }^{n}} = cos\,n\theta +i\,sin n\,\theta \] \[{{\beta }^{n}} = cos n\theta - i sin n\theta \] [Using De Moivre Theorem] \[{{\alpha }^{n}}+\,\,{{\beta }^{n}}=\,\,2cos\,n\theta \,\,and\,\,{{\alpha }^{n}}\cdot \,{{\beta }^{n}}=1\] \[\therefore \] The required equation is \[{{x}^{2}}-2x\cos n\,\theta +1=0\]
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