Select write the most appropriate answer from the given alternative in each following questions:-
The principal solution of tan $x = - \sqrt{3}$ are:

\frac{- π}{3} a n d \frac{2 π}{3}

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\frac{π}{3} a n d \frac{5 π}{3}

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\frac{2 π}{3} a n d \frac{5 π}{3}

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\frac{2 π}{3} a n d \frac{4 π}{3}

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Solution

The correct option is A $\frac{- π}{3} a n d \frac{2 π}{3}$
$tan x = - \sqrt{3} = tan (- \frac{π}{3})$
$g e n e r a l s o l u t i o n$
$x = n π - \frac{π}{3}$
$p r i n c i p a l s o l u t i o n$
$f o r n = 0$
$x = - \frac{π}{3}$
$f o r n = 1$
$x = \frac{2 π}{3}$
$H e n c e p r i n c i p a l s o l u t i o n i s : - \frac{π}{3} a n d \frac{2 π}{3}$

Suggest Corrections

Similar questions

x

(x + \frac{π}{3}) + tan (x + \frac{2 π}{3}) = 3 ⟹ tan 3 x = . . . . . .

2 f (s i n x) + f (c o s x) = x \forall x ϵ

tan θ + tan [θ + \frac{π}{3}] + tan [θ + \frac{2 π}{3}] = 3

2^{1 + | cos x | + {cos}^{2} x + | cos x |^{3} + . . . \infty} = 4

s i n^{- 1} (s i n \frac{2 π}{3}) = \frac{2 π}{3}

t a n^{- 1} (t a n \frac{2 π}{3}) = \frac{- π}{3}

c o s^{- 1} (c o s \frac{2 π}{3}) = \frac{2 π}{3}

s e c^{- 1} (s e c \frac{5 π}{4}) = \frac{3 π}{4}

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