Solve the Following Equation: 2 Sin 2 X + √ 3 Cos X + 1 = 0

प्रश्न

Solve the following equation:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

बेरीज

उत्तर

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]
\[ \Rightarrow 2 - 2 \cos^2 x + \sqrt{3} \cos x + 1 = 0\]
\[ \Rightarrow 2 \cos^2 x - \sqrt{3} \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos^2 x - 2\sqrt{3} \cos x + \sqrt{3} \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos x (\cos x - \sqrt{3}) + \sqrt{3} (\cos x - \sqrt{3}) = 0\]
\[ \Rightarrow (2 \cos x + \sqrt{3}) (\cos x - \sqrt{3}) = 0\]

⇒ \[(2 \cos x + \sqrt{3}) = 0\] or

\[(\cos x - \sqrt{3}) = 0\]

\[\cos x = \sqrt{3}\] is not possible.

\[\therefore 2 \cos x + \sqrt{3} = 0 \]
\[ \Rightarrow \cos x = - \frac{\sqrt{3}}{2} \]
\[ \Rightarrow \cos x = \cos \frac{5\pi}{6} \]
\[ \Rightarrow x = 2n\pi \pm \frac{5\pi}{6}, n \in\]

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Trigonometric Equations

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Q 3.3 Q 3.2 Q 3.4

पाठ 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २२]

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आर.डी. शर्मा Mathematics [English] Class 11

पाठ 11 Trigonometric equations
Exercise 11.1 | Q 3.3 | पृष्ठ २२

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Trigonometric Equations

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