Step 1: Understanding the Concept:
We need to find values of \( x \) for which the expression \( f(x) = \sin^2 x + \sin x \cos x \) takes an integer value \( a \). Step 2: Key Formula or Approach:
Rewrite \( f(x) \) using double angles:
\( f(x) = \frac{1 - \cos 2x}{2} + \frac{\sin 2x}{2} = \frac{1}{2} + \frac{1}{2}(\sin 2x - \cos 2x) \).
Using \( A \sin \theta + B \cos \theta = \sqrt{A^2 + B^2} \sin(\theta + \phi) \):
\( f(x) = \frac{1}{2} + \frac{\sqrt{2}}{2} \sin(2x - \pi/4) \). Step 3: Detailed Explanation:
The range of \( \sin(2x - \pi/4) \) is \( [-1, 1] \).
So, range of \( f(x) \) is \( [\frac{1}{2} - \frac{1}{\sqrt{2}}, \frac{1}{2} + \frac{1}{\sqrt{2}}] \).
Approximate values: \( 1/\sqrt{2} \approx 0.707 \), so range is \( [-0.207, 1.207] \).
Possible integers \( a \) are \( 0 \) and \( 1 \). Case 1: \( a = 0 \)
\( \sin x(\sin x + \cos x) = 0 \)
\( \sin x = 0 \implies x = -\pi, 0, \pi \) (3 solutions).
\( \sin x + \cos x = 0 \implies \tan x = -1 \implies x = -\pi/4, 3\pi/4 \) (2 solutions).
Total for \( a = 0 \) is 5. Case 2: \( a = 1 \)
\( \sin^2 x + \sin x \cos x = 1 \implies \sin x \cos x = 1 - \sin^2 x = \cos^2 x \)
\( \cos x (\sin x - \cos x) = 0 \)
\( \cos x = 0 \implies x = -\pi/2, \pi/2 \) (2 solutions).
\( \tan x = 1 \implies x = \pi/4, -3\pi/4 \) (2 solutions).
Total for \( a = 1 \) is 4.
Total elements = \( 5 + 4 = 9 \). Step 4: Final Answer:
The total number of elements in set \( S \) is 9.
