Question

The number of possible real solution(ss) of \(y\) in the equation \(y^2 - 2y\cos x + 1 = 0\) is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Divide the equation by y and recall that \(y+\dfrac{1}{y}\) never lies strictly between \(-2\) and \(2\) for a real, nonzero y.

Answer 1

The Correct Option is C

Step 1: Treat the equation as a quadratic in y.
The equation \(y^2-2y\cos x+1=0\) is a quadratic in \(y\) with \(a=1\), \(b=-2\cos x\), \(c=1\). Its discriminant is \[ \Delta = b^2-4ac = 4\cos^2 x - 4 \]

Step 2: Find when y can be real.
For \(y\) to be real, we need \(\Delta\geq0\), i.e. \(4\cos^2x-4\geq0\), i.e. \(\cos^2x\geq1\). But for any real angle \(x\), \(\cos x\) always lies between \(-1\) and \(1\), so \(\cos^2x\) can never exceed \(1\); it can at most equal \(1\). So the only way to satisfy \(\cos^2x\geq1\) is \(\cos^2x=1\) exactly, which forces \(\Delta=0\), not \(\Delta>0\).

Step 3: Solve \(\cos^2x=1\).
\(\cos^2x=1\) means \(\cos x=1\) or \(\cos x=-1\), i.e. \(x=0^{\circ}\) (or any multiple of \(360^{\circ}\)) or \(x=180^{\circ}\) (or any odd multiple of \(180^{\circ}\)).

Step 4: Find the value of y at each case.
Since \(\Delta=0\) here, the quadratic has one repeated root \(y=\dfrac{-b}{2a}=\cos x\). So when \(\cos x=1\), \(y=1\); when \(\cos x=-1\), \(y=-1\). These are the only two real values \(y\) can ever take across all valid \(x\); for every other value of \(x\), \(\Delta<0\) and \(y\) is not real at all.

Final Answer:
The equation has exactly \(2\) possible real values of \(y\), namely \(1\) and \(-1\). \[ \boxed{2} \]