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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-

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Question Number 32682 by rahul 19 last updated on 31/Mar/18
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 ?
$$\boldsymbol{{I}}{f}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}\:} \:{are}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${acos}\:\mathrm{2}{x}+{bsin}\:{x}\:=\:{c}\:{and}\: \\ $$$$\mathrm{2}{sin}\:{x}_{\mathrm{1}} {sinx}_{\mathrm{2}} =\:{sin}\:{x}_{\mathrm{1}} +{sinx}_{\mathrm{2}} .\:\boldsymbol{{T}}{hen}\: \\ $$$${the}\:{value}\:{of}\:\:\frac{{b}}{{c}−{a}}\:{is}\:? \\ $$
Answered by MJS last updated on 31/Mar/18
cos 2x=2cos^2 x−1=1−2sin^2 x acos 2x+bsin x−c=0 −2asin^2 x+bsin x+a−c=0 sin^2 x−(b/(2a))sin x−((a−c)/(2a))=0 sin x=(b/(4a))±((√(8a^2 −8ac+b^2 ))/(4a))=p±q 2(p−q)(p+q)=(p−q)+(p+q) 2p^2 −2q^2 =2p 2((b/(4a)))^2 −2(((√(8a^2 −8ac+b^2 ))/(4a)))^2 =2(b/(4a)) (b^2 /(8a^2 ))−((8a^2 −8ac+b^2 )/(8a^2 ))=(b/(2a)) ((c−a)/a)=(b/(2a)) ((c−a)/b)=(1/2) (b/(c−a))=2
$$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} \:{x}−\mathrm{1}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x} \\ $$$${a}\mathrm{cos}\:\mathrm{2}{x}+{b}\mathrm{sin}\:{x}−{c}=\mathrm{0} \\ $$$$−\mathrm{2}{a}\mathrm{sin}^{\mathrm{2}} \:{x}+{b}\mathrm{sin}\:{x}+{a}−{c}=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}−\frac{{b}}{\mathrm{2}{a}}\mathrm{sin}\:{x}−\frac{{a}−{c}}{\mathrm{2}{a}}=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}=\frac{{b}}{\mathrm{4}{a}}\pm\frac{\sqrt{\mathrm{8}{a}^{\mathrm{2}} −\mathrm{8}{ac}+{b}^{\mathrm{2}} }}{\mathrm{4}{a}}={p}\pm{q} \\ $$$$\mathrm{2}\left({p}−{q}\right)\left({p}+{q}\right)=\left({p}−{q}\right)+\left({p}+{q}\right) \\ $$$$\mathrm{2}{p}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} =\mathrm{2}{p} \\ $$$$\mathrm{2}\left(\frac{{b}}{\mathrm{4}{a}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\sqrt{\mathrm{8}{a}^{\mathrm{2}} −\mathrm{8}{ac}+{b}^{\mathrm{2}} }}{\mathrm{4}{a}}\right)^{\mathrm{2}} =\mathrm{2}\frac{{b}}{\mathrm{4}{a}} \\ $$$$\frac{{b}^{\mathrm{2}} }{\mathrm{8}{a}^{\mathrm{2}} }−\frac{\mathrm{8}{a}^{\mathrm{2}} −\mathrm{8}{ac}+{b}^{\mathrm{2}} }{\mathrm{8}{a}^{\mathrm{2}} }=\frac{{b}}{\mathrm{2}{a}} \\ $$$$\frac{{c}−{a}}{{a}}=\frac{{b}}{\mathrm{2}{a}} \\ $$$$\frac{{c}−{a}}{{b}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{b}}{{c}−{a}}=\mathrm{2} \\ $$
Commented by rahul 19 last updated on 31/Mar/18
you have assumed (p−q)=sin x_1 and (p+q)=sin x_2 ,right ? Thank u sir!
$${you}\:{have}\:{assumed}\:\left({p}−{q}\right)={sin}\:{x}_{\mathrm{1}} \\ $$$${and}\:\left({p}+{q}\right)={sin}\:{x}_{\mathrm{2}} \:,{right}\:? \\ $$$${Thank}\:{u}\:{sir}! \\ $$
Commented by MJS last updated on 31/Mar/18
yes
$$\mathrm{yes} \\ $$

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