• # question_answer The number of real values of $a$ satisfying the equation ${{a}^{2}}-2a\sin x+1=0$ is A) Zero B) One C) Two D) Infinite

Correct Answer: C

Solution :

Given equation  ${{a}^{2}}-2a\sin x+1=0$ $\therefore$ $a=\frac{2\sin x\pm \sqrt{4{{\sin }^{2}}x-4}}{2}$ $=\sin x\pm \sqrt{-(1-{{\sin }^{2}}x)}$ $a=\sin x\pm i\cos x$ If $x=\frac{\pi }{2}$Þ $a=1,$$x={{270}^{o}}$Þ $a=-1$.

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