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The General Solution of the Equation 7 Cos 2 X + 3 Sin 2 X = 4 is

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प्रश्न

The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is

विकल्प

  • \[x = 2 n\pi \pm \frac{\pi}{6}, n \in Z\]

     

  • \[x = 2 n\pi \pm \frac{2\pi}{3}, n \in Z\]

     

  • \[x = n\pi \pm \frac{\pi}{3}, n \in Z\]
  • none of these

MCQ
योग
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उत्तर

\[x = n\pi \pm \frac{\pi}{3}, n \in Z\]
Given:
\[7 \cos^2 x + 3 \sin^2 x = 4\]
\[ \Rightarrow 7 \cos^2 x + 3 (1 - \cos^2 x) = 4\]
\[ \Rightarrow 7 \cos^2 x + 3 - 3 \cos^2 x = 4\]
\[ \Rightarrow 4 \cos^2 x + 3 = 4\]
\[ \Rightarrow 4 (1 - \cos^2 x) = 3\]
\[ \Rightarrow 4 \sin^2 x = 3\]
\[ \Rightarrow \sin^2 x = \frac{3}{4}\]
\[ \Rightarrow \sin x = \frac{\sqrt{3}}{2}\]
\[ \Rightarrow \sin x = \sin \frac{\pi}{3}\]
\[ \Rightarrow x = n\pi \pm \frac{\pi}{3}, n \in Z\]
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अध्याय 11: Trigonometric equations - Exercise 11.3 [पृष्ठ २७]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 11 Trigonometric equations
Exercise 11.3 | Q 5 | पृष्ठ २७

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