A set of values of theta for which the system | vedclass

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

Question

(D) For a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$. $\left|\begin{array}{ccc} \sin 3 	heta & -1

Vedclass · Accepted Answer

$n \pi+(-1)^n \frac{\pi}{6}$ (here $n$ is any integer)

Vedclass · Answer

(D) For a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$. $\left|\begin{array}{ccc} \sin 3 	heta & -1 & 1 \ \cos 2 	heta & 4 & 3 \ 2 & 7 & 7 \end{array}ight| = 0$ Expanding along the first row: $\sin 3 	heta (28 - 21) - (-1) (7 \cos 2 	heta - 6) + 1 (7 \cos 2 	heta - 8) = 0$ $7 \sin 3 	heta + 7 \cos 2 	heta - 6 + 7 \cos 2 	heta - 8 = 0$ $7 \sin 3 	heta + 14 \cos 2 	heta - 14 = 0$ Dividing by $7$: $\sin 3 	heta + 2 \cos 2 	heta - 2 = 0$ Using $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$ and $\cos 2 	heta = 1 - 2 \sin^2 	heta$: $(3 \sin 	heta - 4 \sin^3 	heta) + 2(1 - 2 \sin^2 	heta) - 2 = 0$ $3 \sin 	heta - 4 \sin^3 	heta + 2 - 4 \sin^2 	heta - 2 = 0$ $-4 \sin^3 	heta - 4 \sin^2 	heta + 3 \sin 	heta = 0$ $-\sin 	heta (4 \sin^2 	heta + 4 \sin 	heta - 3) = 0$ $-\sin 	heta (2 \sin 	heta - 1)(2 \sin 	heta + 3) = 0$ Since $2 \sin 	heta + 3 
eq 0$ for any real $	heta$,we have: $\sin 	heta = 0 \Rightarrow 	heta = n \pi$ $\sin 	heta = \frac{1}{2} \Rightarrow 	heta = n \pi + (-1)^n \frac{\pi}{6}$ Thus,the set of values is $	heta = n \pi$ or $	heta = n \pi + (-1)^n \frac{\pi}{6}$.

$A$ set of values of $\theta$ for which the system of equations $(\sin 3 \theta) x-y+z=0$,$(\cos 2 \theta) x+4 y+3 z=0, 2 x+7 y+7 z=0$ has non-trivial solutions,is

Similar Questions

If $x^{3}-2x^{2}-9x+18=0$ and $A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$,then the maximum value of $A$ is

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

If $\begin{bmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{bmatrix}$ is a singular matrix,then the value of $x$ is:

If $\left|\begin{array}{ccc}x & 4 & 6 \\ 2 & 3 & -9 \\ 5 & 6 & 1\end{array}\right|+\left|\begin{array}{ccc}5 & 6 & 1 \\ 6 & 4 & 5 \\ 2 & 3 & -9\end{array}\right|=\left|\begin{array}{ccc}2 & 3 & -9 \\ 1-2 x & -8 & -11 \\ 5 & 6 & 1\end{array}\right|$,then $x=$ . . . . . .

If points $(5, 5)$,$(10, k)$ and $(-5, 1)$ are collinear,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module