if equation sin x + sec x −2tan x −1 = 0 has roots x_1 & x_2 , then the possible value of sin x_1 −cos x_2 ? (a) 4/5 (b) 3/4 (c) 4/3 (d) 3/2 (e) 2


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Question Number 87419 by john santu last updated on 04/Apr/20

$if equation \sin x + \sec x - 2 \tan x - 1 = 0$ $has roots x_{1} & x_{2}, then the$ $possible value of \sin x_{1} - \cos x_{2} ?$ $(a) 4 / 5 (b) 3 / 4 (c) 4 / 3$ $(d) 3 / 2 (e) 2$
Commented by john santu last updated on 04/Apr/20

$how sir your solution$
Commented by MJS last updated on 04/Apr/20

$none of these$ $for 0 ⩽ x < 2 π I get x = 0 \lor x \approx 2.44853$
Commented by john santu last updated on 04/Apr/20

$my solution$ $let \tan x = t$ $\frac{t}{\sqrt{1 + t^{2}}} + \sqrt{1 + t^{2}} - 2 t - 1 = 0$ $\frac{t^{2} + t + 1}{\sqrt{1 + t^{2}}} = 2 t + 1$ $t^{2} + t + 1 = (2 t + 1) \sqrt{1 + t^{2}}$
Commented by john santu last updated on 04/Apr/20
$if cos x_1 +cos x_2 = 2. correct mr Mjs ? from t_1 = 0 we get x = 2πn { ((x_1 =0)),((x_2 = 2π)) :}$
$if \cos x_{1} + \cos x_{2} = 2 .$ $correct mr Mjs ?$ $from t_{1} = 0$ $we get x = 2 π n {\begin{cases} x_{1} = 0 \\ x_{2} = 2 π \end{cases}$
Commented by MJS last updated on 04/Apr/20

$this is somehow true but the given equation$ $has infinite solutions and {it}^{'} s asked for 2$ $solutions x_{1} and x_{2} . I think x_{1} \neq x_{2} \land x_{1} \neq x_{2} + 2 n π$ $but anyway \sin x_{1} - \cos x_{2} \neq any of the$ $given options$ $the question is very strange . . .$
Commented by john santu last updated on 04/Apr/20

$yes sir i agree$
	Answered by MJS last updated on 04/Apr/20

	$t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t$ $\frac{- 2 t (t^{3} - 3 t^{2} + t - 1)}{t^{4} - 1} = 0$ $t_{1} = 0$ $t^{3} - 3 t^{2} + t - 1 = 0$ $t = z + 1$ $z^{3} - 2 z - 2 = 0$ $D = \frac{{(- 2)}^{3}}{27} + \frac{{(- 2)}^{2}}{4} = \frac{19}{27} > 0 \Rightarrow Cardano$ $z = \sqrt[3]{1 + \frac{\sqrt{57}}{9}} + \sqrt[3]{1 - \frac{\sqrt{57}}{9}}$ $\Rightarrow t_{2} = 1 + \sqrt[3]{1 + \frac{\sqrt{57}}{9}} + \sqrt[3]{1 - \frac{\sqrt{57}}{9}} \approx 2.76929$ $\Rightarrow$ $x_{1} = 0$ $x_{2} = 2 \arctan (1 + \sqrt[3]{1 + \frac{\sqrt{57}}{9}} + \sqrt[3]{1 - \frac{\sqrt{57}}{9}}) \approx 2.44853$
	Commented by john santu last updated on 04/Apr/20

	$i^{'} m stuck in Cardano sir$
	Commented by MJS last updated on 04/Apr/20
	$x^3 +ax^2 +bx+c=0 let x=t−(a/3) t^3 −((a^2 −3b)/3)t+((2a^3 −9ab+27c)/(27))=0 −((a^2 −3b)/3)=p∧((2a^3 −9ab+27c)/(27))=q t^3 +pt+q=0 D=(p^3 /(27))+(q^2 /4) { ((>0 ⇒ Cardano)),((<0 ⇒ trigonimetric solution)) :} Cardano u=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) v=((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) [we need the real roots i.e. ((−1))^(1/3) =−1] t_1 =u+v t_2 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v t_3 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v trigononetric solution t_(j+1) =((2(√(−3p)))/3)sin (((2jπ)/3)−(1/3)arcsin ((3(√3)q)/(2p(√(−p))))) with j=0, 1, 2$
	$x^{3} + {a x}^{2} + b x + c = 0$ $let x = t - \frac{a}{3}$ $t^{3} - \frac{a^{2} - 3 b}{3} t + \frac{2 a^{3} - 9 a b + 27 c}{27} = 0$ $- \frac{a^{2} - 3 b}{3} = p \land \frac{2 a^{3} - 9 a b + 27 c}{27} = q$ $t^{3} + p t + q = 0$ $D = \frac{p^{3}}{27} + \frac{q^{2}}{4} {\begin{cases} > 0 \Rightarrow Cardano \\ < 0 \Rightarrow trigonimetric solution \end{cases}$ $Cardano$ $u = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} v = \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}}$ $[we need the real roots i . e . \sqrt[3]{- 1} = - 1]$ $t_{1} = u + v$ $t_{2} = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v$ $t_{3} = (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v$ $trigononetric solution$ $t_{j + 1} = \frac{2 \sqrt{- 3 p}}{3} \sin (\frac{2 j π}{3} - \frac{1}{3} \arcsin \frac{3 \sqrt{3} q}{2 p \sqrt{- p}})$ $with j = 0, 1, 2$
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