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 $a\cos x+b\sin x=c$

Ecuatii trigonometrice de forma acos x+bsin x=c

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