The system of equations: 2 x cos 2 θ+y sin 2 θ-2 sin θ=0 x sin 2 θ+2 y sin 2 θ=-2 cos θ x sin θ-y cos θ=0, for all values of θ, can

Question

The system of equations: 2 x cos 2 θ+y sin 2 θ-2 sin θ=0 x sin 2 θ+2 y sin 2 θ=-2 cos θ x sin θ-y cos θ=0, for all values of θ, can (A) have a unique non - trivial solution (B) not have a solution (C) have infinite solutions (D) have a trivial solution

Tardigrade Admin · Accepted Answer

slope of (1) and (2) is cot θ ⇒(1) and (2) are parallel and slope of (3) is tan θ ⇒ no solution. Using R2 arrow R2-(2 cos θ) R3 and R1 arrow R1+(2 sin θ) R3, the value of determinat is 4

Q. The system of equations :
$2 x cos^{2} θ + y sin 2 θ - 2 sin θ = 0$
$x sin 2 θ + 2 y sin^{2} θ = - 2 cos θ$
$x sin θ - y cos θ = 0$ , for all values of $θ$ , can

have a unique non - trivial solution

not have a solution

have infinite solutions

have a trivial solution

slope of (1) and (2) is $cot θ \Rightarrow (1)$ and (2) are parallel and slope of (3) is $tan θ$
$\Rightarrow$ no solution.
Using $R_{2} \to R_{2} - (2 cos θ) R_{3}$ and $R_{1} \to R_{1} + (2 sin θ) R_{3}$ , the value of determinat is 4

Q. The system of equations : 2xcos2θ+ysin2θ−2sinθ=0 xsin2θ+2ysin2θ=−2cosθ xsinθ−ycosθ=0, for all values of θ, can