The number of values of theta in (0, pi) for | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.The coefficient matrix is:$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.The coefficient matrix is:$A = \begin{bmatrix} 1 & 3 & 7 \ -1 & 4 & 7 \ \sin 3	heta & \cos 2	heta & 2 \end{bmatrix}$Setting the determinant to zero:$\begin{vmatrix} 1 & 3 & 7 \ -1 & 4 & 7 \ \sin 3	heta & \cos 2	heta & 2 \end{vmatrix} = 0$Expanding along the first row:$1(8 - 7\cos 2	heta) - 3(-2 - 7\sin 3	heta) + 7(-\cos 2	heta - 4\sin 3	heta) = 0$$8 - 7\cos 2	heta + 6 + 21\sin 3	heta - 7\cos 2	heta - 28\sin 3	heta = 0$$14 - 14\cos 2	heta - 7\sin 3	heta = 0$$2 - 2\cos 2	heta - \sin 3	heta = 0$Using $\cos 2	heta = 1 - 2\sin^2 	heta$ and $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$:$2 - 2(1 - 2\sin^2 	heta) - (3\sin 	heta - 4\sin^3 	heta) = 0$$2 - 2 + 4\sin^2 	heta - 3\sin 	heta + 4\sin^3 	heta = 0$$4\sin^3 	heta + 4\sin^2 	heta - 3\sin 	heta = 0$$\sin 	heta (4\sin^2 	heta + 4\sin 	heta - 3) = 0$$\sin 	heta (2\sin 	heta - 1)(2\sin 	heta + 3) = 0$Since $	heta \in (0, \pi)$,$\sin 	heta > 0$. Thus,$\sin 	heta = 1/2$ is the only valid solution.$\sin 	heta = 1/2 \implies 	heta = \pi/6, 5\pi/6$.There are $2$ such values.

The number of values of $\theta \in (0, \pi)$ for which the system of linear equations $x + 3y + 7z = 0$,$-x + 4y + 7z = 0$,and $(\sin 3\theta)x + (\cos 2\theta)y + 2z = 0$ has a non-trivial solution is:

Similar Questions

The roots of the equation $\left| \begin{matrix} x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x \end{matrix} \right| = 0$ are equal to

If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is

If the vertices of a triangle are $(-2, 6)$,$(3, -6)$,and $(1, 5)$,then the area of the triangle is

$\left|\begin{array}{ccc} \log e & \log e^2 & \log e^3 \\ \log e^2 & \log e^3 & \log e^4 \\ \log e^3 & \log e^4 & \log e^5 \end{array}\right| \text{ is equal to: }$

The roots of the determinant equation (in $x$) $\left| \begin{array}{ccc} a & a & x \\ m & m & m \\ b & x & b \end{array} \right| = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module