Advertisement Remove all ads

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

Options

  • 0

  • 5

  • 6

  • 10

Advertisement Remove all ads

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 \[\left[ 0, \pi \right], \left[ 2\pi, 3\pi \right] and \left[ 4\pi, 5\pi \right] .\]
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Q 21 | Page 28
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×