Ecuatii trigonometrice

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Ecuatiile trigonometrice fundamentale
Fie a un numar real. Ecuatiile in necunoscuta x
\sin x=a, x\in R
\cos x=a,x\in R
\tan x=a, x\in\left\{\left(2k+1\right)\frac{\pi}{2}| k\in Z\right\}
se numesc ecuatiile trigonomerice fundamentale. In legatura cu ecuatiile de mai sus se pun doua probleme:
– existenta solutiei: are ecuatia cel putin o solutie?
– multimea solutiilor: daca ecuatia are solutie, care sunt toate solutiile ecuatiei?
Ecuatia \sin x=a
Conditia de existenta a solutiei ecuatiei este a\in\left[-1,1\right] sau |a|\leq 1
Daca a\in\left(-\infty,-1\right)\cup\left(1,+\infty\right), adica |a|>1 atunci ecuatia nu are solutie.
Daca |a|\leq 1 stim ca ecuatia \sin x=a are solutia unica in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] si anume \arcsin a.
Propozitie: Daca |a|\leq 1 atunci multimea solutiilor ecuatiei \sin x=a este \left\{\left(-1\right)^{n}\arcsin a+n\pi|n\in Z\right\}
Daca |a|>1 ecuatia nu are solutie.
Propozitie:
\sin x=1\Leftrightarrow x=\frac{\pi}{2}+2n\pi, n\in Z
\sin x=0\Leftrightarrow x=n\pi, n\in Z
\sin x=-1\Leftrightarrow x=-\frac{\pi}{2}+2n\pi, n\in Z
Ecuatia \cos x=a
Propozitie:
Daca |a|\leq 1, atunci multimea solutiilor ecuatie \cos x=a este
\left\{\pm\arccos x+2n\pi|n\in Z\right\}
Daca |a|>1, ecuatia nu are solutie.
Propozitie:
\cos x=1\Leftrightarrow x=2n\pi, \in Z
\cos x=0\Leftrightarrow x=\left(2n+1\right)\frac{\pi}{2}, n\in Z
\cos x=-1\Rightarrow x=\pi+2n\pi,n\in Z.
Ecuatii trigonometrice care se reduc la ecuatii fundamentale.
Nu exista o metoda generala pentru a rezolva ecuatiile trigonomtrice. Dar exista anumite procedee particulare prin care anumite ecuatii se reduc la ecuatii funamentale.
Ecuatii de forma \sin u\left(x\right)=\sin v\left(x\right) sau \cos u\left(x\right)=\cos v\left(x\right)
Exemplu:
\sin 5x=\sin 7x\Rightarrow \sin 5x-\sin 7x=0\Rightarrow 2\sin\frac{5x-7x}{2}\cdot\cos\frac{5x+7x}{2}=0\Rightarrow 2\sin\frac{-2x}{2}\cdot\cos\frac{12x}{2}=0\Rightarrow 2\sin\left(-x\right)\cdot\cos6x=0
Astfel avem ca
\sin\left(-x\right)=0\Rightarrow -\sin x=0\Rightarrow \sin x=0\Rightarrow x=k\pi
Sau
\cos 6x=0\Rightarrow 6x=\left(2n+1\right)\frac{\pi}{2}\Rightarrow x=\left(2n+1\right)\frac{\pi}{12}, k,n\in Z
Deci multimea solutiilor ecuatiei este:
\left\{k\pi|k\in Z\right\}\cup\left\{\left(2n+1\right)\frac{\pi}{12}|n\in Z\right\}
b) \cos 10x=\cos 5x\Rightarrow \cos 10x-\cos5x=0\Rightarrow
-2\sin\frac{10x+5x}{2}\cdot\sin\frac{10x-5x}{2}=0\Rightarrow -2\sin\frac{15x}{2}\cdot\sin\frac{5x}{2}=0\Rightarrow \sin\frac{15x}{2}\cdot\sin\frac{5x}{2}=0
Astfel, fie
\sin\frac{15x}{2}=0\Rightarrow \frac{15x}{2}=k\pi\Rightarrow x=\frac{2}{15}\cdot k\pi\Rightarrow x=\frac{2k\pi}{15}
Sau
\sin\frac{5x}{2}=0\Rightarrow\frac{5x}{2}=n\pi\Rightarrow 5x=2n\pi\Rightarrow x=\frac{2n\pi}{5}, k,n\in Z
Astfel multimea solutiilor ecuatiei este:
\left\{\frac{2k\pi}{15}|k\in Z\right\}\cup\left\{2n\frac{\pi}{5}|n\in Z\right\}=\left\{2n\frac{\pi}{5}|n\in Z\right\} (deoarece prima multime este inclusa in cea de-a doua multime.)
c) \sin 6x=\cos 4x
Pentru a rezolva o ecuatie de acest tip procedam astfel:
\sin u\left(x\right)=\cos v\left(x\right)\Leftrightarrow \sin u\left(x\right)=\sin\left(\frac{\pi}{2}-v\left(x\right)\right)
Sau
\sin u\left(x\right)=\cos v\left(x\right)\Leftrightarrow \cos\left(\frac{\pi}{2}-u\left(x\right)\right)=\cos v\left(x\right)
Astfel pentru ecuatia de mai sus avem
\sin 6x=\cos 4x\Leftrightarrow \sin 6x=\sin\left(\frac{\pi}{2}-4x\right)\Leftrightarrow \sin 6x-\sin\left(\frac{\pi}{2}-4x\right)=0\Leftrightarrow 2\sin\frac{6x-\frac{\pi}{2}+4x}{2}\cdot\cos\frac{6x+\frac{\pi}{2}-4x}{2}=0\Leftrightarrow 2\sin\frac{10x-\frac{\pi}{2}}{2}\cdot\cos\frac{2x+\frac{\pi}{2}}{2}=0\Leftrightarrow \sin\frac{10x-\frac{\pi}{2}}{2}\cdot\cos\frac{2x+\frac{\pi}{2}}{2}=0
Astfel avem ca:
\sin\frac{10x-\frac{\pi}{2}}{2}=0\Leftrightarrow \sin\left(5x-\frac{\pi}{4}\right)=0\Leftrightarrow 5x-\frac{\pi}{4}=k\pi\Rightarrow 5x=k\pi+\frac{\pi}{4}\Leftrightarrow 5x=\frac{4k\pi+\pi}{4}\Leftrightarrow x=\frac{4k\pi+\pi}{20}\Leftrightarrow x=\frac{4k\pi}{20}+\frac{\pi}{20}=k\frac{\pi}{5}+\frac{\pi}{20}
Sau
\cos\frac{2x+\frac{\pi}{2}}{2}=0\Leftrightarrow \cos\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=\left(2n+1\right)\frac{\pi}{2}\Leftrightarrow x=\left(2n+1\right)\frac{\pi}{2}-\frac{\pi}{4}\Leftrightarrow x=2n\cdot\frac{\pi}{2}+\frac{\pi}{2}-\frac{\pi}{4}\Leftrightarrow x=n\pi+\frac{2\cdot \pi-\pi}{4}\Leftrightarrow x=n\pi+\frac{\pi}{4}
Deci multimea solutiilor ecuatiei este:
\left\{\frac{\pi}{20}+k\frac{\pi}{5}| k\in Z\right\}\cup\left\{\frac{\pi}{4}+n\pi|n\in Z\right\}
Ecuatii trigonometrice care se reduc la ecuatii algebrice
Consideram ecuatiile, unde a,b,c\in R, a\neq 0
a\sin^{2}x+b\sin x+c=0
a\cos^{2} x+b\cos x+c=0
Prin efectuarea unor substitutii, fiecare din ecuatiile de mai sus se reduc la ecuatii de gradul al doilea.
Exemplu:
Sa se rezolve ecuatiile:
a) \sin^{2}x+\sin x-6=0
Notamc \sin x=t
astfel ecuatia de mai sus devine:
t^{2}+t-6=0
Astfel am obtinut o ecuatie de gradul al doilea pe care o rezolvam:
\Delta=1^{2}-4\cdot 1\cdot\left(-6\right)=1+24=25
t_{1}=\frac{-1+\sqrt{25}}{2\cdot 1}=\frac{-1+5}{2}=\frac{4}{2}=2
Dar si
t_{2}=\frac{-1-\sqrt{25}}{2\cdot 1}=\frac{-1-5}{2}=\frac{-6}{2}=-3
Astfel avem ca
\sin x=2 sau \sin x=-3
Astfel observam ca 2\notin\left[-1,1\right], dar si -3\notin\left[-1,1\right]
Deci ecuatia nu are solutii.
b)4\cos^{2}x-4\cos x+1=0
astfel notam
\cos x=t si ecuatia devine:
4t^{2}-4t+1=0
Calculam
\Delta=\left(-4\right)^{2}-4\cdot 4\cdot 1=16-16=0
Deci solutia ecuatiei este
t=\frac{-b}{2\cdot a}=\frac{-\left(-4\right)}{2\cdot 4}=\frac{4}{8}=\frac{1}{2}
Astfel avem ca
\cos x=\frac{1}{2}\Leftrightarrow x=\pm\frac{\pi}{3}+2n\pi, n\in Z

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