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General Solution of Trigonometric Equation

The smallest ...

Question

$The smallest positive angle which satisfies the quation 2 s i n^{2} θ + \sqrt{3} c o s θ + 1 = 0 i s$

$\frac{5 π}{6}$

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$\frac{2 π}{3}$

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$\frac{π}{3}$

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$\frac{π}{6}$

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Solution

The correct option is A
$\frac{5 π}{6}$

$2 s i n^{2} θ + \sqrt{3} c o s θ + 1 = 0 \Rightarrow 2 (1 - c o s^{2} θ) + \sqrt{3} c o s θ + 1 = 0 \Rightarrow 2 - 2 c o s^{2} θ + \sqrt{3} c o s θ + 1 = 0 \Rightarrow 2 c o s^{2} θ - \sqrt{3} c o s θ - 3 \Rightarrow 2 c o s^{2} θ - 2 \sqrt{3} c o s θ + \sqrt{3} c o s t θ - 3 = 0 \Rightarrow 2 c o s θ (c o s θ - \sqrt{3}) + \sqrt{3} (c o s θ - \sqrt{3}) = 0 \Rightarrow (2 c o s θ + \sqrt{3}) (c o s θ - \sqrt{3}) = 0 \Rightarrow 2 c o s θ + \sqrt{3} = 0 o r ‘, c o s θ - \sqrt{3} = 0 ∴ c o s θ = - \frac{\sqrt{3}}{2} o r c o s θ = \sqrt{3} i s n o t p o s s i b l e . \Rightarrow c o s θ = c o s (\frac{5 π}{6}) \Rightarrow θ = 2 n π \pm \frac{5 π}{6}, n \in Z F o r n = 0, t h e v a l u e o f θ i s \pm \frac{5 π}{6} H e n c e, t h e s m a l l e s t p o s i t i v e a n g l e i s \frac{5 π}{6} .$

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2 \sin^{2} θ + \sqrt{3} \cos θ + 1 = 0

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2 c o s^{2} θ + \sqrt{3} s i n θ + 1 = 0

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2 {cos}^{2} θ + \sqrt{3} sin θ + 1 = 0

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