Question

Let $x \in [-\pi, \pi]$ & $S = \{ x : \sin x (\sin x + \cos x) = a, a \in I \}$, then number of elements in set $S$ is equal to :

• A. 5
• B. 10
• C. 9
• D. 4

Answer 1

The Correct Option is C

Step 1: Simplify the trigonometric expression.
Let $f(x) = \sin x (\sin x + \cos x) = \sin^2 x + \sin x \cos x$.
Using trigonometric identities:
$\sin^2 x = \frac{1 - \cos 2x}{2}$
$\sin x \cos x = \frac{\sin 2x}{2}$
So, $f(x) = \frac{1 - \cos 2x + \sin 2x}{2} = \frac{1}{2} + \frac{1}{2}(\sin 2x - \cos 2x)$.


Step 2: Find the range of $f(x)$.
The range of $(\sin \theta - \cos \theta)$ is $[-\sqrt{2}, \sqrt{2}]$.
Therefore, the range of $f(x)$ is:
$f(x) \in \left[ \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right]$
Approximate values: $\sqrt{2} \approx 1.414$, so:
$f(x) \in [\frac{1 - 1.414}{2}, \frac{1 + 1.414}{2}] \approx [-0.207, 1.207]$.


Step 3: Identify integer values for $a$.
Since $a$ must be an integer ($a \in I$), the possible values for $a$ are $0$ and $1$.


Step 4: Solve for $a = 0$.
$\sin x (\sin x + \cos x) = 0$.
Case 4a: $\sin x = 0 \implies x = -\pi, 0, \pi$ (3 solutions).
Case 4b: $\sin x + \cos x = 0 \implies \tan x = -1 \implies x = -\frac{\pi}{4}, \frac{3\pi}{4}$ (2 solutions).
Total for $a=0$: $3 + 2 = 5$ solutions.


Step 5: Solve for $a = 1$.
$\sin^2 x + \sin x \cos x = 1$
$\sin x \cos x = 1 - \sin^2 x = \cos^2 x$
$\cos x (\sin x - \cos x) = 0$.
Case 5a: $\cos x = 0 \implies x = -\frac{\pi}{2}, \frac{\pi}{2}$ (2 solutions).
Case 5b: $\sin x - \cos x = 0 \implies \tan x = 1 \implies x = -\frac{3\pi}{4}, \frac{\pi}{4}$ (2 solutions).
Total for $a=1$: $2 + 2 = 4$ solutions.


Step 6: Total number of solutions.
Total solutions = $5 + 4 = 9$.
The number of elements in set $S$ is 9. (Option 3)