If - 3 cos x -1-sin x then x-

The correct option is B 2nπ+π/2, 2nπ π/6 √( 3 )cos x =1 sin x √( 3 ) #160;/ 2 cos x + 1 / 2 sin x = 1 / 2 ⇒cos (x π #160;/ 6 ) =co ...

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Standard XII

Mathematics

Inverse of a Function

If √ 3 cos ...

Question

If $\sqrt{3} cos x = 1 - sin x$ then $x =$

2 n π + \frac{π}{2}, 2 n π - \frac{π}{6}

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n π \pm \frac{π}{2}

n π \pm \frac{π}{3}

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n π \pm \frac{π}{2}, n π - \frac{π}{3}

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n π - \frac{π}{2}, n π \pm \frac{π}{3}

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Solution

The correct option is B $2 n π + \frac{π}{2}, 2 n π - \frac{π}{6}$
$\sqrt{3} cos x = 1 - sin x$
$\frac{\sqrt{3}}{2} cos x + \frac{1}{2} sin x = \frac{1}{2}$
$\Rightarrow cos (x - \frac{π}{6}) = cos \frac{π}{3}$
The general solution is given by
$x - \frac{π}{6} = 2 n π \pm \frac{π}{3}$
$\Rightarrow x = 2 n π + \frac{π}{3} + \frac{π}{6}, x = 2 n π - \frac{π}{3} + \frac{π}{6}$
$\Rightarrow x = 2 n π + \frac{π}{2}, 2 n π - \frac{π}{6}$

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The correct option is B 2nπ+π2,2nπ−π6√3cosx=1−sinx√32cosx+12sinx=12⇒cos(x−π6)=cosπ3The general solution is given by x−π6=2nπ±π3⇒x=2nπ+π3+π6,x=2nπ−π3+π6⇒x=2nπ+π2,2nπ−π6

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