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Find the General Solution of the Following Equation: Sin 2 X = √ 3 2

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प्रश्न

Find the general solution of the following equation:

\[\sin 2x = \frac{\sqrt{3}}{2}\]
योग
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उत्तर

We have:

\[\sin2x = \frac{\sqrt{3}}{2}\]

⇒ \[\sin2x = \sin \frac{\pi}{3}\]

⇒ \[2x = n\pi + ( - 1 )^n \frac{\pi}{3}\]

\[n \in Z\]

⇒ \[x = \frac{n\pi}{2} + ( - 1 )^n \frac{\pi}{6}\],

\[n \in Z\]
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अध्याय 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २१]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 11 Trigonometric equations
Exercise 11.1 | Q 2.01 | पृष्ठ २१

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