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Ecuatii trigonometrice de forma acos x+bsin x=c

Pentru a vedea cum rezolvam acest tip de ecuatii, consideram mai intai ecuatia in necunoscuta x a\cos x+b\sin x=c
unde a,b,c sunt numere reale.

Notam cu S multimea solutiilor ecuatiei.
Astfel distingem mai multe cazuri:

Cazul I. a=0, b=0, astfel ecuatia devine
0\cdot \cos x+0\cdot s\in x=c\Leftrightarrow c=0

Acest caz nu este interesant: daca c=0, avem S=R, iar daca c\neq 0 avem S=\Phi, adica multimea vida.(ecuatia nu are solutii).

Cazul II. Daca a=0, b\neq 0 sau a\neq 0, b=0 astfel daca
a=0, b\neq 0 ecuatia devine b\sin x=c\Rightarrow \sin x=\frac{c}{b}, care are solutie daca si numai daca |c|\leq |b|
Daca a\neq 0, b=0 ecuatia devine a\cos x=c\Rightarrow \cos x=\frac{c}{a}, care are solutie daca si numai daca |c|\leq |a|.

Cazul III
Daca a\neq 0, b\neq 0
In acest caz exista doua metode de rezolvare a ecuatiei, numite: metoda algebrica si metoda unghiului auxiliar.

Metoda algebrica.
Stim ca \cos x, \sin x se pot exprima in functie de \tan \frac{x}{2} daca \frac{x}{2}\neq \left(2k+1\right)\frac{\pi}{2} sau x\neq\left(2k+1\right)\pi, k\in Z
Facand aceasta substitutie.

Stim ca \cos x=\frac{1-tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}
Dar si \sin x=\frac{2\tan^{2}\frac{x}{2}}{1\tan^{2}\frac{x}{2}}
Iar ecuatia de mai sus devine o ecuatie de gradul I sau II in necunoscuta t.
Rezolvarea ecuatiei se desfasoara dupa urmatorul algoritm:

1. Se verifica daca ecuatia are solutii de forma \pi+2k\pi. Acesta revine la testarea egalitatii a=c\Leftrightarrow a+c=0
Astfel daca are loc conditia de mai sus, ecuatia are solutie de forma \pi+2k\pi
Daca egalitatea a+c=0 este adevarata incluziuna
\left\{\pi+2k\pi|k\in Z\right\}\subset S
Iar ecuatia devine
a\cos x+b\sin x=c\Rightarrow a\cos x+b\sin x=-a\Rightarrow
a+a\cos x+b\sin x=0\Rightarrow a\left(1+\cos x\right)+b\sin x=0
\Rightarrow a\cdot 2\cos^{2}\frac{x}{2}+b\sin x=0\Rightarrow 2a\cos^{2}\frac{x}{2}+2b\sin\frac{x}{2}\cos\frac{x}{2}=0
\Rightarrow 2\cos\frac{x}{2}\left(a\cos\frac{x}{2}+b\sin\frac{x}{2}\right)=0

Astfel \cos\frac{x}{2}=0
Sau a\cos\frac{x}{2}+b\sin\frac{x}{2}=0\Leftrightarrow \tan\frac{x}{2}=-\frac{a}{b}
Deci daca a+c=0 solutiile ecuatiei sunt \left\{\pi+2k\pi|k\in Z\right\}\cup\left\{-2\arctan\frac{x}{2}+2m\pi|m\in Z\right\}

Daca a+c\neq 0 face substitutia de mai sus \tan\frac{x}{2}=t
Astfel obtinem o ecuatie de gradul al doliea in necunoscuta t.
Conditia de existenta a solutiilor ecuatiei exprimata unitar
Ecuatia a\cos x+b\sin x=c are solutii daca si numai daca a^{2}+b^{2}\geq c^{2}

Exemple:
Sa se rezolve ecuatiile:
a) \sin x-\cos x =1\Leftrightarrow -\cos x+\sin x=1
Prima data scrim coefienti ecuatiei, astfel avem a=-1, b=1, c=1 si verificam inegaliatea \left(-1\right)+1^{2}\geq 1^{2}\Leftrightarrow 2\geq 1, deci se verifica
Astfel testam egalitatea
a+c=-1+1=0, deci egalitatea este adevarata. Prin urmare ecuatia admite solutii de forma x=\pi+2k\pi
Ecuatia de mai sus devine
-\cos x+\sin x-1=0\Rightarrow
-1\left(1+\cos x\right)+\sin x=0\Leftrightarrow
-1\cdot 2\cos^{2}\frac{x}{2}+\sin x=0\Leftrightarrow
-2\cos^{2}\frac{x}{2}+2\sin\frac{x}{2}\cdot\cos\frac{x}{2}=0\Leftrightarrow
2\cos\frac{x}{2}\left(-\cos\frac{x}{2}+\sin\frac{x}{2}\right)=0

Astfel obtinem ca: \cos\frac{x}{2}=0\Leftrightarrow \frac{x}{2}=\pi+2m\pi,m\in Z(solutia de la primul pas)
Sau -\cos\frac{x}{2}+\sin\frac{x}{2}=0\Rightarrow \tan\frac{x}{2}=-\frac{-1}{1}\Rightarrow \tan\frac{x}{2}=1\Leftrightarrow \frac{x}{2}=\arctan 1\Leftrightarrow x=2\arctan 1+2k\pi\Leftrightarrow x=2\cdot\frac{\pi}{4}+2k\pi\Leftrightarrow x=\frac{\pi}{2}+2k\pi
b)\sin x+7\cos x=7\Leftrightarrow 7\cos x+\sin x=5
Scriem mai intai coeficienti ecuatiei a=7, b=1,c=5

Pasul 1. Egalitatea a+c=7+5=12, deci este falsa. Prin urmare ecuatiaa nu are solutii de forma \pi+2k\pi
Astfel facem substitutia \tan \frac{x}{2}=t
Si ecuatia de mai sus devine: 7\cdot\frac{1-t^{2}}{1+t^{2}}+\frac{2t}{1+t^{2}}=5\Leftrightarrow \frac{7-7t^{2}+2t}{1-t^{2}}=5\Leftrightarrow 7-7t^{2}+2t=5\left(1+t^{2}\right)\Leftrightarrow 7-7t^{2}+2t=5+5t^{2}\Leftrightarrow 7-7t^{2}+2t-5-5t^{2}=0\Leftrightarrow -12t^{2}+2t+2=0
Astfel calculam \Delta=2^{2}-4\cdot\left(-12\right)\cdot 2=4+96=100
Astfel t_{1}=\frac{-2+\sqrt{100}}{2\cdot\left(-12\right)}=\frac{-2+10}{-24}=\frac{8}{-24}=-\frac{1}{3}
Iar t_{2}=\frac{-2-\sqrt{100}}{2\cdot\left(-12\right)}=\frac{-2-10}{-24}=\frac{-12}{-24}=\frac{1}{2}
Astfel stim ca \tan\frac{x}{2}=-\frac{1}{3}\Rightarrow \frac{x}{2}=\arctan\left(-\frac{1}{3}\right)\Rightarrow x=-2\arctan\frac{1}{3}+2n\pi
Dar si \tan\frac{x}{2}=\frac{1}{2}\Rightarrow \frac{x}{2}=\arctan\frac{1}{2}+k\pi\Rightarrow x=2\arctan\frac{1}{2}+2k\pi
Prin urmare S=\left\{2\arctan\frac{1}{2}+2k\pi| k\in Z\right\}\cup\left\{-2\arctan\frac{1}{3}+2n\pi|n\in Z\right\}

In alt articol o sa prezentam si metoda unghiului auxiliar de rezolvare a ecuatiilor trigonometrice de forma a\cos x+b\sin x=c