Complex Numbers And Quadratic Equations question 125
Question: The number of real values of $ a $ satisfying the equation $ a^{2}-2a\sin x+1=0 $ is
Options:
A) Zero
B) One
C) Two
D) Infinite
Show Answer
Answer:
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 $ .
BETA
