If 2 cos θ + sin θ = 1 (θ ≠ (π/2)), then 7 cos θ + 6 sin θ is equal to :

Question

If 2 cos θ + sin θ = 1 (θ ≠ (π/2)), then 7 cos θ + 6 sin θ is equal to : (A) (1/2) (B) 2 (C) (11/2) (D) (46/5)

Tardigrade Admin · Accepted Answer

2 cos Θ=1- sin Θ on squaring both the sides beginarrayl ⇒ 4 cos 2 Θ= sin 2 Θ-2 sin θ+1 ⇒ 4(1- sin 2 Θ)= sin 2 Θ-2 sin Θ+1 ⇒ 5 sin 2 Θ-2 sin Θ-3=0 ⇒ sin Θ=1 ; sin Θ=-(3/5) endarray As per the question θ=(π/2) hence here sin θ=-(3/5) so the cos Θ=(4/5) required answer =7((4/5))+6(-(3/5))=2

Q. If 2 cos $θ$ + sin $θ$ = 1 $(θ \neq = \frac{π}{2}),$ then 7 cos $θ$ + 6 sin $θ$ is equal to :

$\frac{1}{2}$

2

$\frac{11}{2}$

$\frac{46}{5}$

Q. If 2 cos θ + sin θ = 1 (θ=2π​), then 7 cos θ + 6 sin θ is equal to :

21​

2

211​

546​

2cosΘ=1−sinΘ on squaring both the sides ⇒4cos2Θ=sin2Θ−2sinθ+1⇒4(1−sin2Θ)=sin2Θ−2sinΘ+1⇒5sin2Θ−2sinΘ−3=0⇒sinΘ=1;sinΘ=−53​​ As per the question θ=2π​ hence here sinθ=−53​ so the cosΘ=54​ required answer =7(54​)+6(−53​)=2

Q. If 2 cos $θ$ + sin $θ$ = 1 $(θ \neq = \frac{π}{2}),$ then 7 cos $θ$ + 6 sin $θ$ is equal to :

$\frac{1}{2}$

$\frac{11}{2}$

$\frac{46}{5}$