# 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 $\sin^2 x - \cos x = \frac{1}{4}$

• 1

• 2

• 3

• 4

#### 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