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

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

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

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

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

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

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

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

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

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

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

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

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

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