For n- z the general solution of -3 -1 sin- - -3 -1 cos- -2 is n- z

The correct option is A θ =2nπ±π/4+π/12 ( √(3) 1 )sinθ + ( √(3) +1 )cosθ =2 nbsp; nbsp; ....(1)Put √( 3 ) 1=rsinα nbsp; ,√( 3 ) +1= ...

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Standard XII

Mathematics

Cos(A+B)Cos(A-B)

For n∈ z th...

Question

For $n \in z$ the general solution of $(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2$ is $(n ϵ z)$

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 $θ = 2 n π \pm \frac{π}{4} + \frac{π}{12}$
$(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2$ ....(1)
Put $\sqrt{3} - 1 = r sin α, \sqrt{3} + 1 = r cos α$
$\Rightarrow r = 2 \sqrt{2}$
and $tan α = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
$\Rightarrow α = 15^{0} = \frac{π}{12}$
Put these values in (1),
$r cos (θ - α) = 2$
$cos (θ - \frac{π}{12}) = \frac{1}{\sqrt{2}}$
$θ - \frac{π}{12} = 2 n π \pm \frac{π}{4}$
$\Rightarrow θ = 2 n π \pm \frac{π}{4} + \frac{π}{12}$

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For n∈z the general solution of (√3−1)sinθ+(√3+1)cosθ=2 is (nϵz)

The correct option is A θ=2nπ±π4+π12(√3−1)sinθ+(√3+1)cosθ=2 ....(1)Put √3−1=rsinα,√3+1=rcosα⇒r=2√2and tanα=√3−1√3+1⇒α=150=π12Put these values in (1),rcos(θ−α)=2cos(θ−π12)=1√2θ−π12=2nπ±π4⇒θ=2nπ±π4+π12

For $n \in z$ the general solution of $(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2$ is $(n ϵ z)$