Question

The general solution of the equation \(2\sin ^{2}x+\sin ^{2}2x=2\) is \(x=(2n+1)\alpha\) or \(x=m\pi \pm \beta\), \(m,n\in \mathbb{Z}\), then \(\alpha\) and \(\beta\) are the roots of the quadratic equation _______

• A. \(8x^{2}-6\pi x+\pi ^{2}=0\)
• B. \(4x^{2}-6\pi x+\pi ^{2}=0\)
• C. \(8x^{2}+6\pi x+\pi ^{2}=0\)
• D. \(4x^{2}+6\pi x+\pi ^{2}=0\)

Hint:

Replace sin 2x by 2 sin x cos x and factor the equation to find the two families of solutions.

Answer 1

The Correct Option is A

Step 1: Rewrite
\(\sin^{2}2x=4\sin^{2}x\cos^{2}x\), so \(2\sin^{2}x+4\sin^{2}x\cos^{2}x=2\). Divide by \(2\): \(\sin^{2}x+2\sin^{2}x\cos^{2}x=1\).

Step 2: Factor
\(2\sin^{2}x\cos^{2}x=1-\sin^{2}x=\cos^{2}x\), so \(\cos^{2}x(2\sin^{2}x-1)=0\).

Step 3: Solve
\(\cos x=0\) gives \(x=(2n+1)\dfrac{\pi}{2}\), so \(\alpha=\dfrac{\pi}{2}\).
\(\sin^{2}x=\dfrac{1}{2}\) gives \(x=m\pi\pm\dfrac{\pi}{4}\), so \(\beta=\dfrac{\pi}{4}\).

Step 4: Form the Quadratic
Sum of roots \(=\dfrac{3\pi}{4}\), product \(=\dfrac{\pi^{2}}{8}\). The equation is \(x^{2}-\dfrac{3\pi}{4}x+\dfrac{\pi^{2}}{8}=0\), that is \(8x^{2}-6\pi x+\pi^{2}=0\), option (A).

Final Answer:
The quadratic is \(8x^{2}-6\pi x+\pi^{2}=0\). \[ \boxed{\text{(A) }8x^{2}-6\pi x+\pi^{2}=0} \]