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Trigonometrical Equations
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The number of values of $x$ in the interval $[0, 5\pi]$ satisfying the equation $3sin^2x\, \,-\,\, 7sinx + 2 = 0$ is
A
$0$
B
$5$
C
$6$
D
$10$
Solution
$3 \sin ^{2} x-7 \sin x+2=0$
$\Rightarrow \quad(3 \sin x-1)(\sin x-2)=0$
$\Rightarrow \quad 3 \sin x=1$ or $\quad \sin x=2$
$\Rightarrow \quad \sin x=\frac{1}{3} \quad[\because \sin x=2 \text { is not possible }]$
since $x \in[0,5 \pi]$
$\therefore $ $6$ values of $x$ will be possible.
[$\because x$ will lie in ${I^{st}}{\rm{ }}or{\rm{ }}I{I^{nd}}$ quadrant]
Standard 11
Mathematics
