If 3cosθ+sinθ=2, then θ=

3cosθ+sinθ=2 ⇒32cosθ+12sinθ=12 ⇒cos(θ−π6)=cosπ4 ⇒θ−π6=2nπ±π4,n∈z ⇒θ=2nπ±π4+π6,n∈z

If 3cosθ+sinθ=2, then θ=

3cosθ+sinθ=2 ⇒32cosθ+12sinθ=12 ⇒cos(θ−π6)=cosπ4 ⇒θ−π6=2nπ±π4,n∈z ⇒θ=2nπ±π4+π6,n∈z

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\sqrt{3} \cos θ + \sin θ = \sqrt{2},$ then $θ =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 n π \pm \frac{π}{6}; n \in Z$

$n π + {(- 1)}^{n} \frac{π}{4} + \frac{π}{6}; n \in Z$

$n π + \frac{π}{6}; n \in Z$

$2 n π \pm \frac{π}{4} + \frac{π}{6}; n \in Z$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\sqrt{3} \cos θ + \sin θ = \sqrt{2} \Rightarrow \frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ = \frac{1}{\sqrt{2}} \Rightarrow \cos (θ - \frac{π}{6}) = \cos \frac{π}{4} \Rightarrow θ - \frac{π}{6} = 2 n π \pm \frac{π}{4}, n \in z \Rightarrow θ = 2 n π \pm \frac{π}{4} + \frac{π}{6}, n \in z$