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-2-Then-the-value-of-b-c-a-is-

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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