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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The number of values of $x$ in the interval $[0,3\pi]$ .Satisfying the equation $2\sin^2x+5\sin x-3=0$ is


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Since $2\sin^2x+5\sin x-3=0$
$(2\sin x-1)(\sin x+3)=0$
$2\sin x-1=0$
$2\sin x=1$
$\sin x=\large\frac{1}{2}$
$\sin x+3=0$
$\sin x \neq-3$
Hence the intersection point is $y=\large\frac{1}{2}$ and $y=\sin x$ is $4$
Hence the no of solutions in $[0,3\pi]$ is 4.
Hence (d) is the correct option.
answered Nov 18, 2013 by sreemathi.v

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