Solve the following trigonometric equation :
$tan 2 x = \sqrt{3}$

\frac{n π}{2} + \frac{π}{6}, n \in Z

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n π + \frac{π}{6}, n \in Z

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\frac{n π}{3} - \frac{π}{12}, n \in Z

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None of these

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Solution

The correct option is A $\frac{n π}{2} + \frac{π}{6}, n \in Z$
$tan 2 x = \sqrt{3}$
$tan 2 x = tan \frac{π}{3}$

$2 x = n π + \frac{π}{3}, n \in Z$

$x = \frac{n π}{2} + \frac{π}{6}, n \in Z$

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Solve the following trigonometric equation :tan2x=√3

The correct option is A nπ2+π6,n∈Ztan2x=√3tan2x=tanπ32x=nπ+π3,n∈Zx=nπ2+π6,n∈Z

Solve the following trigonometric equation :
$tan 2 x = \sqrt{3}$

The correct option is A $\frac{n π}{2} + \frac{π}{6}, n \in Z$
$tan 2 x = \sqrt{3}$
$tan 2 x = tan \frac{π}{3}$

$2 x = n π + \frac{π}{3}, n \in Z$

$x = \frac{n π}{2} + \frac{π}{6}, n \in Z$