Question

Consider the system of linear equations in x, y and z ;
(sin 3$\theta$) x - y + z = 0 ......(i)
(cos 2$\theta$)x + 4y + 3z = 0 ......(ii)
2x + 7y+ 7z = 0 ......(iii)

The value of $\theta$ for which the system has nontrivial solution is

• A. θ = nπ, nπ + (-1)ⁿ (π/6), where n ϵ I.
• B. θ= nπ+(π/6),, nπ + (-1)ⁿ (π/4), where n ϵ I.
• C. θ= nπ-π/6, nπ + (-1)ⁿ (π/3), where n ϵ I.
• D. θ=2nπ , where n ϵ I.

Answer 1

The correct option is A

θ = nπ, nπ + (-1)ⁿ (π/6), where n ϵ I.

Eliminating x, y, z from the system of linear equations(i) , (ii), (iii), then

$\left|\begin{array}{ccc}sin3\theta & -1& 1\\ cos2\theta & 4& 3\\ 2& 7& 7\end{array}\right|$ =0
$or(28-21)sin3\theta -(-7-7)cos2\theta +2(-3-4)=0or,7sin3\theta +14cos2\theta -14=0or,sin3\theta +cos2\theta -2=0or,(3sin\theta -4si{n}^3\theta )+2(1-2si{n}^2\theta )-2=0or4si{n}^3\theta +4si{n}^2\theta -3sin\theta =0orsin\theta (4si{n}^2\theta +4sin\theta -3)=0orsin\theta (2sin\theta -1)(2sin\theta +3)=0$
Either sin $\theta$ = 0 or sin $\theta$ =$\frac{1}{2}$
or $sin\theta =-\frac{3}{2}$ but sin$\theta$ = 0,
or $sin\theta =\frac{1}{2}$ is possible or $sin\theta =-\frac{3}{2}$ is not possible Now, sin$\theta$ = 0
or, $\theta$ = n$\pi$ where n $\in$ I.
And sin $\theta$ =$\frac{1}{2}$ = $sin\left(\frac{\pi }{6}\right)$,
or, $\theta$ = n$\pi$ +$(-1{)}^n\frac{\pi }{6}$, where n $\in$ I.
Hence the required values of $\pi$ are $\theta =n\pi$,
n$\pi$ + $(-1{)}^n(\frac{\pi }{6})$, where n $\in$ I.