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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 = 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 \dots

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} + {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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