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(A)$(\dfrac{\pi }{6},\dfrac{{5\pi }}{6})$

(B)$(0,\dfrac{\pi }{6}) \cup (\dfrac{{5\pi }}{6},\pi )$

(C)$(0,\dfrac{{5\pi }}{6})$

(D)$(0,\pi )$

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$2{x^2} - 4x + 2\sin \theta - 1 = 0$

We are solving this question by using general equation, that is $a{x^2} + bx + c = 0$ if we suppose it`s root is $\alpha ,\beta $ and it`s of opposite sign that`s mean $\alpha .\beta < 0$. In these type of question the value of $\alpha ,\beta $ low root get value of $\dfrac{c}{a} < 0$

comparing $2{x^2} - 4x + 2\sin \theta - 1 = 0$ with general $a{x^2} + bx + c = 0$

a = 2 b = -4 c = $2\sin \theta $

In this equation both roots are of opposite sing so we apply $\dfrac{c}{a} < 0$ (same equation) $\dfrac{{2\sin \theta - 1}}{2} < 0$

$2\sin \theta - 1 < 0$

$\operatorname{Sin} \theta < \dfrac{1}{2}$

$D > 0$ [ where D = ${b^2} - 4ac$ ]

Value of D will be ${4^2} - 2.(2\sin \theta - 1) > 0$

= $16 - 4.2.(2\sin \theta - 1) > 0$

= $16 - 8(2\sin \theta - 1) > 0$

Now open the brackets and multiply by 8

=\[16 - 16\sin \theta + 8 > 0\]

Add the numbers which are 16+8

=$24 - 16\sin \theta > 0$

compare the equation

= $24 > 16\sin \theta $

= $\operatorname{Sin} \theta < \dfrac{{24}}{{16}}$

This quantity is greater than 1. This is true for all values of $\theta $ because the value of $\sin \theta $ is less than 1.

$\operatorname{Sin} \theta $ should be less than $\dfrac{3}{2} = 1.5$

$\operatorname{Sin} \theta > 0.5$

$\theta $=$(0,\dfrac{\pi }{6}),(\dfrac{{5\pi }}{6},\pi )$

*Both roots are positive (If a and b are opposite in sign and a and c are same in sign)

*Both roots are negative (If a,b,c are all of same sign)

*Roots are of opposite sign (If a and c are of opposite sign)

*roots equal but opposite in sing (If b=0)