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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

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