Question

Solve the equation for $x$.
$\left|\begin{array}{ccc}{a}^2& a& 1\\ \sin (n+1)x& \sin nx& \sin (n-1)x\\ \cos (n+1)x& \cos nx& \cos (n-1)x\end{array}\right|=0$. Given that $a>0$

• A. $x=n\pi$
• B. $x=(n-1)\pi$
• C. $x=(n+1)\pi$
• D. None of these

Answer 1

The correct option is A $x=n\pi$

$\left|\begin{array}{ccc}{a}^2& a& 1\\ \sin (n+1)x& \sin nx& \sin (n-1)x\\ \cos (n+1)x& \cos nx& \cos (n-1)x\end{array}\right|=0$

$\Rightarrow a^2[\sin nx.\cos (n-1)x-\cos nx.\sin (n-1)x]+a\left[\sin \right(n-1)x.\cos (n+1)x-\cos (n-1)x.\sin (n+1\left)x\right]+1\left[\sin \right(n+1)x.\cos nx-\cos (n+1)x.\sin nx]=0$
$\Rightarrow a^2\sin x-a\sin 2x+\sin x=0\Rightarrow \sin x(a^2+1-2a\cos x)=0\Rightarrow \sin x=0or\cos x=\frac{a^2+1}{2a}\Rightarrow \sin x=0or\cos x=1(\because a^2+1\ge 2a,a>0\Rightarrow \frac{a^2+1}{2a}\ge 1)\Rightarrow x=n\pi orx=2n\pi ,n\in I\Rightarrow x=n\pi$