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The Number of Values of ​X in [0, 2π] that Satisfy the Equation Sin 2 X − Cos X = 1 4 - Mathematics

MCQ
Sum

The number of values of ​x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]

Options

  • 1

  • 2

  • 3

  • 4

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Solution

2
\[\sin^2 x - \cos x = \frac{1}{4}\]
\[ \Rightarrow (1 - \cos^2 x) - \cos x = \frac{1}{4}\]
\[ \Rightarrow 4 - 4 \cos^2 x - 4 \cos x = 1\]
\[ \Rightarrow 4 \cos^2 x + 4 \cos x - 3 = 0\]
\[ \Rightarrow 4 \cos^2 x + 6 \cos x - 2 \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos x ( 2 \cos x + 3) - 1 ( 2 \cos x + 3) = 0\]
\[ \Rightarrow (2 \cos x + 3 ) (2 \cos x - 1) = 0\]
\[\Rightarrow 2 \cos x + 3 = 0\] or, \[2 \cos x - 1 = 0\]
\[\Rightarrow \cos x = - \frac{3}{2}\] or \[\cos x = \frac{1}{2}\]
Here,
\[\cos x = - \frac{3}{2}\]  is not possible.
\[\cos x = \frac{1}{2}\]
\[\Rightarrow \cos x = \cos \frac{\pi}{3}\]
\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}\]
Now for n = 0 and 1, the values of \[x are \frac{\pi}{3}, \frac{5\pi}{3}\text{ and }\frac{7\pi}{3},\text{ but }\frac{7\pi}{3} \text{ is not in }\] \[\left[ 0, 2\pi \right]\]
Hence, there are two solutions in \[\left[ 0, 2\pi \right]\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Q 14 | Page 27
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