1-cosx-1-sinx-8-find-the-value-of-x-

Question Number 72492 by petrochengula last updated on 29/Oct/19

\frac{1}{cosx} + \frac{1}{sinx} = 8

find the value of x

Commented by petrochengula last updated on 29/Oct/19

help please

Answered by behi83417@gmail.com last updated on 29/Oct/19

sinx + cosx = 8 sinx . cosx

{(sinx + cosx)}^{2} = {(8 sinx . cosx)}^{2}

1 + 2 sinx . cosx = {64 \sin}^{2} {xcos}^{2} x [let : sinxcosx = y]

{65 y}^{2} = y^{2} + 2 y + 1 = {(y + 1)}^{2}

\Rightarrow \sqrt{65} y = \pm (2 y + 1) \Rightarrow {\begin{cases} y = \frac{1}{\sqrt{65} - 2} \\ y = \frac{- 1}{\sqrt{65} + 2} \end{cases}

1) sinx . cosx = \frac{1}{\sqrt{65} - 2} \Rightarrow \sin 2 x = \frac{2}{\sqrt{65} - 2}

\Rightarrow x = k π \pm \frac{1}{2} \sin^{- 1} [\frac{2}{\sqrt{65} - 2}] (k \in z)

2) sinx . cosx = \frac{- 1}{\sqrt{65} + 2} \Rightarrow \sin 2 x = \frac{- 2}{\sqrt{65} + 2}

\Rightarrow x = l π \pm \frac{1}{2} \sin^{- 1} [\frac{- 2}{\sqrt{65} + 2}] (l \in z)

Commented by petrochengula last updated on 29/Oct/19

thanks

Answered by Tanmay chaudhury last updated on 29/Oct/19

s i n x + c o s x = 4 (2 s i n x c o s x)

\sqrt{1 + s i n 2 x} = 4 s i n 2 x

1 + a = 16 a^{2} \to 16 a^{2} - a - 1 = 0

a = \frac{1 \pm \sqrt{1 - 4.16 . (- 1)}}{2 \times 16} = \frac{1 \pm \sqrt{65}}{32}

\sqrt{65} \approx 8.06

a = \frac{9.06}{32}, \frac{- 7.06}{32}

a \approx 0.28, - 0.22

s i n 2 x = 0.28 = s i n (0.284) [n o t e a n g l e i n r a d i a n

2 x = 0.284 \to x = 0.142

t o o k h e l p o f c a l c u l a t o r

Commented by petrochengula last updated on 29/Oct/19

thanks

Commented by Tanmay chaudhury last updated on 29/Oct/19

m o s t w e l c o m e \dots

Answered by MJS last updated on 29/Oct/19

x = 2 \arctan t \Rightarrow \cos x = \frac{1 - t^{2}}{1 + t^{2}}; \sin x = \frac{2 t}{1 + t^{2}}

\frac{1 + t^{2}}{1 - t^{2}} + \frac{1 + t^{2}}{2 t} = 8

transforming \Rightarrow

t^{4} - 18 t^{3} + 14 t - 1 = 0

we can exactly solve this

(t^{2} - (9 + \sqrt{65}) t - (8 + \sqrt{65})) (t^{2} - (9 - \sqrt{65}) t - (8 r - \sqrt{65})) = 0

t_{1, 2} = \frac{9 + \sqrt{65}}{2} \pm \frac{\sqrt{178 + 22 \sqrt{65}}}{2}

t_{3, 4} = \frac{9 - \sqrt{65}}{2} \pm \frac{\sqrt{178 - 22 \sqrt{65}}}{2}

x_{i} = 2 π n + 2 \arctan t with n \in Z

\Rightarrow

x_{1} \approx - 1.45953 + 2 π n

x_{2} \approx 3.03033 + 2 π n

x_{3} \approx . 143562 + 2 π n

x_{4} \approx 1.42723 + 2 π n

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