Question

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

विकल्प

• 0

• 5

• 6

• 10

Answer 1

 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 \[\left[ 0, \pi \right], \left[ 2\pi, 3\pi \right] and \left[ 4\pi, 5\pi \right] .\]