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$T h e n u m b e r o f v a l u e s o f x i n t h e i n t e r v a l [0, 3 π] s a t i s f y i n g t h e e q u a t i o n 2 \sin^{2} x + 5 \sin x - 3 = 0 i s$

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Correct option is B

Given equation is 2sin2x+5sinx−3=0⇒sinx+32sinx−1=0⇒ sinx=12∵sinx+3≠0Since x∈o,3π, ⇒x=π6,5π6,13π6,17π6

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The number of values of x in the interval 0,3π satisfying the equation 2sin2x+5sinx−3=0 is

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$T h e n u m b e r o f v a l u e s o f x i n t h e i n t e r v a l [0, 3 π] s a t i s f y i n g t h e e q u a t i o n 2 \sin^{2} x + 5 \sin x - 3 = 0 i s$