Question

Let \[ S=\{x\in[-\pi,\pi]:\sin x(\sin x+\cos x)=a,\; a\in\mathbb{Z}\}. \] Then \(n(S)\) is equal to:

• A. \(3\)
• B. \(6\)
• C. \(7\)
• D. \(9\)

Answer 1

The Correct Option is B

Concept: We analyze the range of the expression \[ f(x)=\sin x(\sin x+\cos x) \] and determine how many integer values it can take.
Step 1:Expand the expression.} \[ f(x)=\sin^2 x+\sin x\cos x \] Using identities, \[ \sin^2x=\frac{1-\cos2x}{2}, \qquad \sin x\cos x=\frac{\sin2x}{2} \] Thus \[ f(x)=\frac{1}{2}-\frac{\cos2x}{2}+\frac{\sin2x}{2} \] \[ f(x)=\frac12+\frac{\sin2x-\cos2x}{2} \]
Step 2:Find the range.} \[ \sin2x-\cos2x=\sqrt2\sin\left(2x-\frac{\pi}{4}\right) \] Hence \[ f(x)=\frac12+\frac{\sqrt2}{2}\sin\left(2x-\frac{\pi}{4}\right) \] Thus \[ -\frac{\sqrt2}{2} \le \frac{\sqrt2}{2}\sin(\cdot) \le \frac{\sqrt2}{2} \] So \[ \frac12-\frac{\sqrt2}{2} \le f(x) \le \frac12+\frac{\sqrt2}{2} \] \[ -0.207 \le f(x) \le 1.207 \]
Step 3:Find integer values in this interval.} Possible integers: \[ 0,\;1 \]
Step 4:Solve each case.} For \(f(x)=0\): \[ \sin x(\sin x+\cos x)=0 \] Solutions in \([-\pi,\pi]\): \[ x=-\pi,0,\pi,-\frac{\pi}{4},\frac{3\pi}{4} \] For \(f(x)=1\): \[ \sin x(\sin x+\cos x)=1 \] This gives one additional solution. Total solutions \(=6\). Hence \[ n(S)=6 \]