The quadratic equation (1 - sin θ )x2+2(1 - sin ⁡ θ )x-3sin ⁡ θ =0 has both roots complex for all θ lying in the interval

Question

The quadratic equation (1 - sin θ )x2+2(1 - sin ⁡ θ )x-3sin ⁡ θ =0 has both roots complex for all θ lying in the interval (A) (- (π /2) , (π /2)) (B) (0 , (3 π /2)) (C) ((π /6) , (7 π /6)) (D) ((7 π /6) , (11 π /6))

Tardigrade Admin · Accepted Answer

D=4(1 - sin θ )2+4(1 - sin â¡ θ )⋅ 3sin â¡ θ < 0 ⇒ 4(1 - sin θ )(1 - sin â¡ θ + 3 sin â¡ θ ) < 0 ⇒ 4(1 - sin θ )(1 + 2 sin â¡ θ ) < 0 ⇒ 8(sin θ - 1)(sin â¡ θ + (1/2))>0 ⇒ sin θ < -(1/2 )or sin â¡ θ >1 sin θ < -(1/2)⇒ θ ∈ ((7 π /6) , (11 π /6))

Q. The quadratic equation $(1 - s in θ) x^{2} + 2 (1 - s in θ) x - 3 s in θ = 0$ has both roots complex for all $θ$ lying in the interval

$(- \frac{π}{2}, \frac{π}{2})$

$(0, \frac{3 π}{2})$

$(\frac{π}{6}, \frac{7 π}{6})$

$(\frac{7 π}{6}, \frac{11 π}{6})$

Q. The quadratic equation (1−sinθ)x2+2(1−sin⁡θ)x−3sin⁡θ=0 has both roots complex for all θ lying in the interval

(−2π​,2π​)

(0,23π​)

(6π​,67π​)

(67π​,611π​)

D=4(1−sinθ)2+4(1−sin⁡θ)⋅3sin⁡θ<0 ⇒ 4(1−sinθ)(1−sin⁡θ+3sin⁡θ)<0 ⇒ 4(1−sinθ)(1+2sin⁡θ)<0 ⇒ 8(sinθ−1)(sin⁡θ+21​)>0 ⇒ sinθ<−21​orsin⁡θ>1 sinθ<−21​⇒θ∈(67π​,611π​)

Q. The quadratic equation $(1 - s in θ) x^{2} + 2 (1 - s in θ) x - 3 s in θ = 0$ has both roots complex for all $θ$ lying in the interval

$(- \frac{π}{2}, \frac{π}{2})$

$(0, \frac{3 π}{2})$

$(\frac{π}{6}, \frac{7 π}{6})$

$(\frac{7 π}{6}, \frac{11 π}{6})$