The number of values of x in the interval [0,5π] satisfting the equation 3sin2⁡x−7sin⁡x+2=0 is

A
0
B
5
C
6
D
10

Solution:

$\begin{array}{l} 3 \sin^{2} x - 7 \sin x + 2 = 0 \\ \Rightarrow 3 \sin^{2} x - 6 \sin x - \sin x + 2 = 0 \\ \Rightarrow 3 \sin x (\sin x - 2) - 1 (\sin x - 2) = 0 \\ \Rightarrow (3 \sin x - 1) (\sin x - 2) = 0 \\ \Rightarrow \sin x = \frac{1}{3} or 2 (∵ \sin x \neq 2) \\ \Rightarrow \sin x = \frac{1}{3} (∵ \sin x \neq 2) \end{array}$

Let $\sin^{- 1} \frac{1}{3} = α, 0 < x < \frac{π}{2}$

Then, $α, π - α, 2 π + α, 3 π - α, 4 π + α, 5 π - α$ are the soulution

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

Solution:

Related content