The number of
real solutions of y in equation y

^{2}-2ycosx+1=0 are:
(A)0 (B)1 (C)2 (D)3 (E)
None of these

Solution
follows here:

__Solution:____Funda:__

Roots of the quadratic equation ax

^{2}+bx+c=0 are -b±√(b^{2}-4ac) / 2a
If b

^{2}-4ac ≥ 0, roots are real,
If b

^{2}-4ac < 0, roots are imaginary.
Determinant
= b

^{2}-4ac = (2cosx)^{2}-4 = 4(cos^{2}x-1) = -4sin^{2}x
For
getting real solutions, Determinant ≥ 0 => -4sin

^{2}x ≥ 0
This
is possible only for sin

^{2}x = 0 => cos^{2}x = 1-sin^{2}x = 1 => cosx = ±1
=>
the given equation becomes y

^{2}-2y(±1)+1=0
=>
y

^{2}-2y+1=0 (or) y^{2}+2y+1=0
=>
(y-1)

^{2}= 0 (or) (y+1)^{2}= 0
=>
y = 1 (or) -1

Two
real solutions

**Answer (C)**
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