Resoudre-l-equaation-acos-x-bsin-x-c-a-b-c-R-3-

Question Number 177306 by a.lgnaoui last updated on 03/Oct/22

R e s o u d r e l e q u a a t i o n

a \cos x + b \sin x = c

(a, b, c) \in R^{3}

Answered by Ar Brandon last updated on 03/Oct/22

a \cos x + b \sin x = c \dots eqn (i)

Let R \sin (x + ϑ) = c

\Rightarrow R \sin x \cos ϑ + R \cos x \sin ϑ = c \dots eqn (i i)

Comparing (i) and (i i)

{\begin{cases} R \sin ϑ = a \\ R \cos ϑ = b \end{cases} \Rightarrow \tan ϑ = \frac{a}{b} \Rightarrow ϑ = \tan^{- 1} (\frac{a}{b})

a^{2} + b^{2} = R^{2} \Rightarrow R = \sqrt{a^{2} + b^{2}}

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

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

Commented by a.lgnaoui last updated on 03/Oct/22

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

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

x + α = β \Rightarrow x = β - α

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

Commented by a.lgnaoui last updated on 03/Oct/22

x = \sin^{- 1} (\frac{c}{\sqrt{a^{2} + b^{2}}}) - \tan^{- 1} (\frac{a}{b}) + k 2 π (k \in Z)

Commented by a.lgnaoui last updated on 03/Oct/22

a v e c - 1 ⩽ \frac{c}{\sqrt{a^{2} + b^{2}}} ⩽ 1 (c^{2} ⩽ a^{2} + b^{2})

Answered by a.lgnaoui last updated on 03/Oct/22

o n p e u t p o s e r t = \tan \frac{x}{2}

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

a (1 - t^{2}) + 2 b t = c (1 + t^{2})

t^{2} - \frac{2 b}{a + c} t + \frac{c - a}{c + a} = 0

Δ^{'} = \frac{b^{2}}{{(a + c)}^{2}} - \frac{c - a}{a + c} = \frac{a^{2} + b^{2} - c^{2}}{{(a + c)}^{2}} (c^{2} ⩽ a^{2} + b^{2})

t = \frac{b \pm \sqrt{a^{2} + b^{2} - c^{2}}}{a + c}

\frac{x_{1}}{2} = A rctan (\frac{b + \sqrt{a^{2} + b^{2} - c^{2}}}{a + c}) + (2 k + 1) π

x_{1} = 2 A rctan (\frac{b + \sqrt{a^{2} + b^{2} - c^{2}}}{a + c}) + (2 k + 1) π (a + c \neq 0 e t k \in Z)

x_{2} = 2 A rctan (\frac{b - \sqrt{a^{2} + b^{2} - c^{2}}}{a + c}) + (2 k + 1) π