The number of integral value(s) of $k$ for which ${sin}^{2} x + (cos x - 1) sin x - cos x - k sin x + k = 0$ have exactly three real roots, where $x \in (0, 2 π)$ , is

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Solution

Given equation is
${sin}^{2} x + (cos x - 1) sin x - cos x - k sin x + k = 0 \Rightarrow {sin}^{2} x + sin x \cdot cos x - k sin x - sin x - cos x + k = 0 \Rightarrow sin x (sin x + cos x - k) - 1 (sin x + cos x - k) = 0 \Rightarrow (sin x - 1) (sin x + cos x - k) = 0 \Rightarrow sin x = 1 or sin x + cos x = k$

As $x \in (0, 2 π)$ , then $sin x = 1$ has one real root.
$sin x + cos x = k \Rightarrow sin (x + \frac{π}{4}) = \frac{k}{\sqrt{2}}$
This equation should have two real root, so
$k \in (- \sqrt{2}, \sqrt{2})$
Therefore, the integral value of $k$ are $- 1, 0, 1$
When $k = 1$
$x = \frac{π}{2}, (∵ x \in (0, 2 π))$

Therefore, $k = - 1, 0$
Hence, there are two possible integral values of $k .$

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