Resoudre-dans-R-acos-x-bsin-x-c-x-0-sin-1-sin-x-d-1-d-1-

Question Number 210231 by a.lgnaoui last updated on 03/Aug/24

Resoudre dans R

{\begin{cases} a \cos x - b \sin x = c (x \neq 0) \\ \sin (\frac{1}{\sin x}) = d (- 1 ⩽ d ⩽ + 1) \end{cases}

Commented by mr W last updated on 04/Aug/24

t h e y a r e t w o d i f f e r e n t e q u a t i o n s f o r

v a r i a b l e x . t h e y h a v e d i f f e r e n t r o o t s .

t h e y c a n n o t b e a n e q u a t i o n s y s t e m!

e q n . 1 :

a \cos x - b \sin x = c

\sqrt{a^{2} + b^{2}} \cos (x + \tan^{- 1} \frac{b}{a}) = c

a s s u m e ∣ c ∣⩽ \sqrt{a^{2} + b^{2}}

\cos (x + \tan^{- 1} \frac{b}{a}) = \frac{c}{\sqrt{a^{2} + b^{2}}}

\Rightarrow x + \tan^{- 1} \frac{b}{a} = 2 k π \pm \cos^{- 1} \frac{c}{\sqrt{a^{2} + b^{2}}}

\Rightarrow x = 2 k π - \tan^{- 1} \frac{b}{a} \pm \cos^{- 1} \frac{c}{\sqrt{a^{2} + b^{2}}}

e q n . 2 :

\sin (\frac{1}{\sin x}) = d

\Rightarrow \frac{1}{\sin x} = k π + {(- 1)}^{k} \sin^{- 1} d

\Rightarrow \sin x = \frac{1}{k π + {(- 1)}^{k} \sin^{- 1} d}

\Rightarrow x = n π + {(- 1)}^{n} \sin^{- 1} (\frac{1}{k π + {(- 1)}^{k} \sin^{- 1} d})

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