# The Number of Values of X in the Interval [0, 5 π] Satisfying the Equation 3 Sin 2 X − 7 Sin X + 2 = 0 is - Mathematics

MCQ
Sum

The number of values of x in the interval [0, 5 π] satisfying the equation $3 \sin^2 x - 7 \sin x + 2 = 0$ is

• 0

• 5

• 6

• 10

#### Solution

6
Given:
$3 \sin^2 x - 7 \sin x + 2 = 0$
$\Rightarrow 3 \sin^2 x - 6 \sin x - \sin x + 2 = 0$
$\Rightarrow 3 \sin x (\sin x - 2) - 1 (\sin x - 2) = 0$
$\Rightarrow (3 \sin x - 1) (\sin x - 2) = 0$

$\Rightarrow 3 \sin x - 1 = 0$ or $\sin x - 2 = 0$
Now,"
sin x = 2 is not possible, as the value of sin x  lies between - 1 and 1.
⇒ $\sin x = \frac{1}{3}$
Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval $\left[ 0, \pi \right]$.
Hence, it is positive six times in the interval $\left[ 0, \pi \right]$, viz $\left[ 0, \pi \right], \left[ 2\pi, 3\pi \right] and \left[ 4\pi, 5\pi \right] .$
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Q 21 | Page 28