The general solution of -3-1 sin-3-1 cos-2 iswhere n - Z A- -2 n -4-12B- - n -1 n -2-4-12- -2 n -4D- - n -1 n -4-12

The correct option is A θ = 2 n π±π4+ π12(√(3) 1) sinθ + (√(3)+1) cosθ = 2 ⇒(√(3) 1)2 √(2)sinθ ( √(3)+12 √(2) )cosθ = 1√(2) ⇒sinπ12sinθ + cosπ12cosθ = cosπ4 ⇒c ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

The general s...

Question

The general solution of $(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2$ is
(where $n \in Z$ )

θ = 2 n π \pm \frac{π}{4} + \frac{π}{12}

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θ = n π + (- 1)^{n} \frac{π}{4} + \frac{π}{12}

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θ = 2 n π \pm \frac{π}{4}

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θ = n π + (- 1)^{n} \frac{π}{4} - \frac{π}{12}

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Solution

The correct option is A $θ = 2 n π \pm \frac{π}{4} + \frac{π}{12}$
$(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2$
$\Rightarrow \frac{(\sqrt{3} - 1)}{2 \sqrt{2}} sin θ (\frac{\sqrt{3} + 1}{2 \sqrt{2}}) cos θ = \frac{1}{\sqrt{2}}$
$\Rightarrow sin \frac{π}{12} sin θ + cos \frac{π}{12} cos θ = cos \frac{π}{4}$
$\Rightarrow cos (θ - \frac{π}{12}) = cos \frac{π}{4}$
$\Rightarrow θ - \frac{π}{12} = 2 n π \pm \frac{π}{4}, n \in Z$
So, $θ = 2 n π \pm \frac{π}{4} + \frac{π}{12}, n \in Z$

Suggest Corrections

The general solution of (√3−1)sinθ+(√3+1)cosθ=2 is (where n∈Z)

The correct option is A θ=2nπ±π4+π12(√3−1)sinθ+(√3+1)cosθ=2 ⇒(√3−1)2√2sinθ(√3+12√2)cosθ=1√2 ⇒sinπ12sinθ+cosπ12cosθ=cosπ4 ⇒cos(θ−π12)=cosπ4 ⇒θ−π12=2nπ±π4,n∈Z So, θ=2nπ±π4+π12,n∈Z

The general solution of $(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2$ is
(where $n \in Z$ )