Formule Generale pentru Ecuatii Trigonometrice
Exemplu Rezolvat
Să rezolvăm în \(\mathbb{R}\) ecuația:
\( \sin x + \cos x = 0 \)
1) Determinarea Domeniului de Definiție (D.V.A.)
- Divizăm ecuația prin \(\cos x\):
\( tg x + 1 = 0 \)
- Obținem: \( tg x = -1 \)
2) Soluția Generală
- Folosim formula pentru \(tg x = a\):
\( x = \operatorname{arctg}(-1) + k \pi, k \in \mathbb{Z} \)
- Calculăm: \( \operatorname{arctg}(-1) = -\frac{\pi}{4} \)
3) Soluții în Domeniul D.V.A.
Exerciții
1
Determinați soluțiile reale ale ecuației \(\sqrt{3} \cos x - \sin(2x) = 0\), pentru care \(|x| < 2\).
Răspuns: \( \cos \alpha = 0.6, \ \operatorname{tg} \alpha = \displaystyle \frac{4}{3}, \ \operatorname{ctg} \alpha = \displaystyle \frac{3}{4} \)
2
Determinați soluțiile reale ale ecuației \(\sin(2x) + \cos x = 0\), pentru care \(\cos x > 0\).
Răspuns: \( \sin \alpha = -\displaystyle \frac{7}{25}, \ \operatorname{tg} \alpha = \displaystyle \frac{7}{24}, \ \operatorname{ctg} \alpha = \displaystyle \frac{24}{7} \)
3
Determinați valorile lui \(\beta \in \left(\displaystyle \frac{\pi}{2}, \pi\right)\), pentru care \(tg(\alpha + \beta) = 2\) și \(tg \alpha = -3\).
Răspuns: \( \operatorname{ctg} \alpha = \displaystyle \frac{1}{2}, \ \sin \alpha = -\displaystyle \frac{2\sqrt{5}}{5}, \ \cos \alpha = -\displaystyle \frac{\sqrt{5}}{5} \)
4
\(\text{Rezolvați ecuația} \, \sin 3x = \sin x \, \text{în} \, \mathbb{R}.\)
Răspuns: \( \operatorname{tg} \alpha = -3, \ \sin \alpha = -\displaystyle \frac{1}{\sqrt{10}}, \ \cos \alpha = \displaystyle \frac{3}{\sqrt{10}} \)
5
\(\text{Rezolvați ecuația} \, \cos 2x = \cos^2 x \, \text{în} \, \mathbb{R}.\)
Răspuns: \( \operatorname{tg} \alpha = -\displaystyle \frac{4}{3} \)
6
\(\text{Rezolvați ecuația} \, \operatorname{tg} x = \sqrt{3} \, \text{în intervalul} \, (0, \pi).\)
Răspuns: \( \operatorname{ctg} \alpha = \displaystyle \frac{15}{8} \)
7
\(\text{Rezolvați ecuația} \, \sin x = \cos x \, \text{în} \, \mathbb{R}.\)
Răspuns: \( \cos \alpha = -\displaystyle \frac{12}{13}, \ \operatorname{tg} \alpha = \displaystyle \frac{5}{12}, \ \operatorname{ctg} \alpha = \displaystyle \frac{12}{5} \)
8
\(\text{Fie} \, \cos \alpha = \displaystyle \frac{3}{5}, \, 0 < \alpha < \displaystyle \frac{\pi}{2}. \, \text{Aflați} \, \sin \alpha; \, \operatorname{tg} \alpha; \, \operatorname{ctg} \alpha.\)
Răspuns: \( \sin \alpha = \displaystyle \frac{4}{5}, \ \operatorname{tg} \alpha = \displaystyle \frac{4}{3}, \ \operatorname{ctg} \alpha = \displaystyle \frac{3}{4} \)
9
\(\text{Fie} \, \operatorname{tg} \alpha = -\displaystyle \frac{5}{12}, \, \pi < \alpha < \displaystyle \frac{3\pi}{2}. \, \text{Aflați} \, \sin \alpha; \, \cos \alpha; \, \operatorname{ctg} \alpha.\)
Răspuns: \( \sin \alpha = -\displaystyle \frac{5}{13}, \ \cos \alpha = -\displaystyle \frac{12}{13}, \ \operatorname{ctg} \alpha = -\displaystyle \frac{12}{5} \)
Răspunsuri
1
\( \cos \alpha = 0.6, \ \operatorname{tg} \alpha = \displaystyle \frac{4}{3}, \ \operatorname{ctg} \alpha = \displaystyle \frac{3}{4} \)
2
\( \sin \alpha = -\displaystyle \frac{7}{25}, \ \operatorname{tg} \alpha = \displaystyle \frac{7}{24}, \ \operatorname{ctg} \alpha = \displaystyle \frac{24}{7} \)
3
\( \operatorname{ctg} \alpha = \displaystyle \frac{1}{2}, \ \sin \alpha = -\displaystyle \frac{2\sqrt{5}}{5}, \ \cos \alpha = -\displaystyle \frac{\sqrt{5}}{5} \)
4
\( \operatorname{tg} \alpha = -3, \ \sin \alpha = -\displaystyle \frac{1}{\sqrt{10}}, \ \cos \alpha = \displaystyle \frac{3}{\sqrt{10}} \)
5
\( \operatorname{tg} \alpha = -\displaystyle \frac{4}{3} \)
6
\( \operatorname{ctg} \alpha = \displaystyle \frac{15}{8} \)
7
\( \cos \alpha = -\displaystyle \frac{12}{13}, \ \operatorname{tg} \alpha = \displaystyle \frac{5}{12}, \ \operatorname{ctg} \alpha = \displaystyle \frac{12}{5} \)
8
\( \sin \alpha = \displaystyle \frac{4}{5}, \ \operatorname{tg} \alpha = \displaystyle \frac{4}{3}, \ \operatorname{ctg} \alpha = \displaystyle \frac{3}{4} \)
9
\( \sin \alpha = -\displaystyle \frac{5}{13}, \ \cos \alpha = -\displaystyle \frac{12}{13}, \ \operatorname{ctg} \alpha = -\displaystyle \frac{12}{5} \)
Rezolvări
1
Folosim identitatea fundamentală:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( 0.8^2 + \cos^2 \alpha = 1 \)
\( 0.64 + \cos^2 \alpha = 1 \)
\( \cos^2 \alpha = 0.36 \)
\( \cos \alpha = 0.6 \) (pozitiv în cadranul I)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = \displaystyle \frac{0.8}{0.6} = \displaystyle \frac{4}{3} \)
\( \operatorname{ctg} \alpha = \displaystyle \frac{1}{\operatorname{tg} \alpha} = \displaystyle \frac{3}{4} \)
2
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \sin^2 \alpha + \left(-\displaystyle \frac{24}{25}\right)^2 = 1 \)
\( \sin^2 \alpha + \displaystyle \frac{576}{625} = 1 \)
\( \sin^2 \alpha = 1 - \displaystyle \frac{576}{625} = \displaystyle \frac{49}{625} \)
\( \sin \alpha = -\displaystyle \frac{7}{25} \) (negativ în cadranul III)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = \displaystyle \frac{-\frac{7}{25}}{-\frac{24}{25}} = \displaystyle \frac{7}{24} \)
\( \operatorname{ctg} \alpha = \displaystyle \frac{24}{7} \)
3
Din definiție:
\( \operatorname{ctg} \alpha = \displaystyle \frac{1}{\operatorname{tg} \alpha} = \displaystyle \frac{1}{2} \)
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = 2 \Rightarrow \sin \alpha = 2 \cos \alpha \)
\( (2 \cos \alpha)^2 + \cos^2 \alpha = 1 \)
\( 4 \cos^2 \alpha + \cos^2 \alpha = 1 \)
\( 5 \cos^2 \alpha = 1 \)
\( \cos^2 \alpha = \displaystyle \frac{1}{5} \)
\( \cos \alpha = -\displaystyle \frac{\sqrt{5}}{5} \) (negativ în cadranul III)
\( \sin \alpha = 2 \cdot \left(-\displaystyle \frac{\sqrt{5}}{5}\right) = -\displaystyle \frac{2\sqrt{5}}{5} \)
4
Din definiție:
\( \operatorname{tg} \alpha = \displaystyle \frac{1}{\operatorname{ctg} \alpha} = -3 \)
Folosim identitatea:
\( \sin \alpha = \operatorname{tg} \alpha \cdot \cos \alpha \)
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( (-3 \cos \alpha)^2 + \cos^2 \alpha = 1 \)
\( 9 \cos^2 \alpha + \cos^2 \alpha = 1 \)
\( 10 \cos^2 \alpha = 1 \)
\( \cos^2 \alpha = \displaystyle \frac{1}{10} \)
\( \cos \alpha = \displaystyle \frac{\sqrt{10}}{10} = \displaystyle \frac{1}{\sqrt{10}} \) (pozitiv în cadranul IV)
\( \sin \alpha = -3 \cdot \displaystyle \frac{1}{\sqrt{10}} = -\displaystyle \frac{3}{\sqrt{10}} \)
5
Cadranul II: \(\sin \alpha > 0\), \(\cos \alpha < 0\)
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \left(\displaystyle \frac{4}{5}\right)^2 + \cos^2 \alpha = 1 \)
\( \displaystyle \frac{16}{25} + \cos^2 \alpha = 1 \)
\( \cos^2 \alpha = \displaystyle \frac{9}{25} \)
\( \cos \alpha = -\displaystyle \frac{3}{5} \) (negativ)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = \displaystyle \frac{\frac{4}{5}}{-\frac{3}{5}} = -\displaystyle \frac{4}{3} \)
6
Cadranul III: \(\cos \alpha < 0\), \(\sin \alpha < 0\)
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \sin^2 \alpha + \left(-\displaystyle \frac{8}{17}\right)^2 = 1 \)
\( \sin^2 \alpha + \displaystyle \frac{64}{289} = 1 \)
\( \sin^2 \alpha = \displaystyle \frac{225}{289} \)
\( \sin \alpha = -\displaystyle \frac{15}{17} \) (negativ)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = \displaystyle \frac{-\frac{15}{17}}{-\frac{8}{17}} = \displaystyle \frac{15}{8} \)
\( \operatorname{ctg} \alpha = \displaystyle \frac{8}{15} \)
7
Cadranul III: \(\sin \alpha < 0\), \(\cos \alpha < 0\)
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \left(-\displaystyle \frac{5}{13}\right)^2 + \cos^2 \alpha = 1 \)
\( \displaystyle \frac{25}{169} + \cos^2 \alpha = 1 \)
\( \cos^2 \alpha = \displaystyle \frac{144}{169} \)
\( \cos \alpha = -\displaystyle \frac{12}{13} \) (negativ)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = \displaystyle \frac{-\frac{5}{13}}{-\frac{12}{13}} = \displaystyle \frac{5}{12} \)
\( \operatorname{ctg} \alpha = \displaystyle \frac{12}{5} \)
8
Cadranul I: toate funcțiile pozitive
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \sin^2 \alpha + \left(\displaystyle \frac{3}{5}\right)^2 = 1 \)
\( \sin^2 \alpha + \displaystyle \frac{9}{25} = 1 \)
\( \sin^2 \alpha = \displaystyle \frac{16}{25} \)
\( \sin \alpha = \displaystyle \frac{4}{5} \)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = \displaystyle \frac{4/5}{3/5} = \displaystyle \frac{4}{3} \)
\( \operatorname{ctg} \alpha = \displaystyle \frac{3}{4} \)
9
Cadranul III: \(\sin \alpha < 0\), \(\cos \alpha < 0\)
\( \operatorname{ctg} \alpha = \displaystyle \frac{1}{\operatorname{tg} \alpha} = -\displaystyle \frac{12}{5} \)
Folosim identitatea:
\( \sin^2 \alpha + \cos^2 \alpha = 1 \)
\( \operatorname{tg} \alpha = \displaystyle \frac{\sin \alpha}{\cos \alpha} = -\displaystyle \frac{5}{12} \Rightarrow \sin \alpha = -\displaystyle \frac{5}{12} \cos \alpha \)
\( \left(-\displaystyle \frac{5}{12} \cos \alpha\right)^2 + \cos^2 \alpha = 1 \)
\( \displaystyle \frac{25}{144} \cos^2 \alpha + \cos^2 \alpha = 1 \)
\( \displaystyle \frac{169}{144} \cos^2 \alpha = 1 \)
\( \cos^2 \alpha = \displaystyle \frac{144}{169} \)
\( \cos \alpha = -\displaystyle \frac{12}{13} \)
\( \sin \alpha = -\displaystyle \frac{5}{12} \cdot \left(-\displaystyle \frac{12}{13}\right) = -\displaystyle \frac{5}{13} \)