Question

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

पर्याय

• `1/2`

• 2

• `11/2`

• `46/5`

Answer 1

If 2 cosθ + sinθ = `1(θ ≠ π/2)`, then 7 cosθ + 6 sinθ is equal to `underlinebb(46/5)`.

Explanation:

2 cosθ + sinθ = 1

Squaring both sides, we get

(2 cosθ + sin θ)² = 1²

⇒ 4cos² θ + sin² θ + 4 sinθcosθ = 1

⇒ 3cos² θ + (cos² θ + sin² θ) + 4 sinθcosθ = 1

⇒ 3cos² θ + 1 + 4 sinθcosθ = 1

⇒ 3cos² θ + 4 sinθcosθ = 0

⇒ cosθ(3cosθ + 4sinθ) = 0

⇒ 3cosθ + 4sinθ = 0 

⇒ 3cosθ = -4sinθ 

⇒ `(-3)/4` = tanθ = `sqrt(sec^2theta-1)=(-3)/4`           `(because tanθ = sqrt(sec^2theta-1)=(-3)/4)`

⇒ sec²θ - 1 = `((-3)/4)^2=9/16` 

⇒ sec²θ = `9/16+1=25/16`

⇒ secθ = `5/4`

or cosθ = `4/5`                        ...(i)

Now,  sin² θ + cos² θ = 1

⇒ sin² θ + `(4/5)^2` = 1

⇒ sin² θ = `1-16/25=9/25`

sinθ = ±`3/5`                         ...(ii)

Taking `(sinθ = +3/5)` because `(sinθ = -3/5)` cannot satisfy the given equation.

Therefore; 7 cosθ + 6 sinθ 

= `7xx4/5+6xx3/5=28/5+18/5=46/5`