If x_1 and x_(2 ) are roots of the equation acos 2x+bsin x = c and 2sin x_1 sinx_2 = sin x_1 +sinx


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Question Number 32682 by rahul 19 last updated on 31/Mar/18

$I f x_{1} a n d x_{2} a r e r o o t s o f t h e e q u a t i o n$ $a c o s 2 x + b s i n x = c a n d$ $2 s i n x_{1} {s i n x}_{2} = s i n x_{1} + {s i n x}_{2} . T h e n$ $t h e v a l u e o f \frac{b}{c - a} i s ?$
	Answered by MJS last updated on 31/Mar/18

	$\cos 2 x = {2 \cos}^{2} x - 1 = 1 - {2 \sin}^{2} x$ $a \cos 2 x + b \sin x - c = 0$ $- 2 a \sin^{2} x + b \sin x + a - c = 0$ $\sin^{2} x - \frac{b}{2 a} \sin x - \frac{a - c}{2 a} = 0$ $\sin x = \frac{b}{4 a} \pm \frac{\sqrt{8 a^{2} - 8 a c + b^{2}}}{4 a} = p \pm q$ $2 (p - q) (p + q) = (p - q) + (p + q)$ $2 p^{2} - 2 q^{2} = 2 p$ $2 {(\frac{b}{4 a})}^{2} - 2 {(\frac{\sqrt{8 a^{2} - 8 a c + b^{2}}}{4 a})}^{2} = 2 \frac{b}{4 a}$ $\frac{b^{2}}{8 a^{2}} - \frac{8 a^{2} - 8 a c + b^{2}}{8 a^{2}} = \frac{b}{2 a}$ $\frac{c - a}{a} = \frac{b}{2 a}$ $\frac{c - a}{b} = \frac{1}{2}$ $\frac{b}{c - a} = 2$
	Commented by rahul 19 last updated on 31/Mar/18

	$y o u h a v e a s s u m e d (p - q) = s i n x_{1}$ $a n d (p + q) = s i n x_{2}, r i g h t ?$ $T h a n k u s i r!$
	Commented by MJS last updated on 31/Mar/18

	$yes$
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