Question

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

• A. $x=\frac{r\pi }{2},r\in I$
• B. $x=\frac{r\pi }{4},r\in I$
• C. $x=\frac{(2r+1)\pi }{2},r\in I$
• D. $x=r\pi ,r\in I$

Answer 1

Answer: (D)

$x=r\pi ,r\in I$

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

For this a row should be zero

$\sin mx$ is zero for m to be integral multiplies of $\pi$

$\Rightarrow x=r\pi rϵI$

$\cos mx$ can also be equal to zero but the generalized solution depends on whether m is even or odd

Hence the condition for which det is zero for all values of n is $x=r\pi ,rϵI$