Найдите корни уравнений, принадлежащие данному промежутку:[latex]tgx=1, \\ x(\frac{\pi}{2} ; \frac{3\pi}{2} ) \\ \\ cosx=- \frac{ \sqrt{3} }{2}, \\ x[- \pi ;\pi] [/latex]Во втором уравнении по формуле косинуса плюс-минус arccos...

[latex]\mathrm{tg}x=1 \\\ x= \frac{ \pi }{4} + \pi n, \ n\in Z \\\ \frac{ \pi }{2}\ \textless \ \frac{ \pi }{4} + \pi n\ \textless \ \frac{ 3\pi }{2} \\\ \frac{ 1 }{2}\ \textless \ \frac{1 }{4} + n\ \textless \ \frac{ 3 }{2} \\\ \frac{ 1 }{2}-\frac{1 }{4}\ \textless \ n\ \textless \ \frac{ 3 }{2}-\frac{1 }{4} \\\ \frac{1 }{4}\ \textless \ n\ \textless \ \frac{ 5 }{4} \\\ n=1: \ x=\frac{ \pi }{4} + \pi =\frac{ 5\pi }{4}[/latex] Ответ: 5π/4 [latex]\cos x=- \frac{\sqrt{3}}{2} \\\ x=\pm \frac{5 \pi }{6}+2\pi n, \ n \in Z \\\ \left[\begin{array}$-\pi \leq \frac{5\pi}{6}+2\pi k\leq\pi \\-\pi \leq-\frac{5\pi}{6}+2\pi m\leq\pi \end{array}\right. \\\ \left[\begin{array}$-1\leq\frac{5}{6}+2k\leq1\\-1\leq-\frac{5}{6}+2m \leq1\end{array}\right. \\\ \left[\begin{array}$-1-\frac{5 }{6}\leq2k\leq1-\frac{5}{6}\\-1+\frac{5}{6}\leq2m\leq1+\frac{5}{6} \end{array}\right.[/latex] [latex]\left[\begin{array}$-\frac{11}{6}\leq2k\leq\frac{1}{6}\\-\frac{1}{6}\leq2m\leq\frac{11}{6} \end{array}\right. \\\ \left[\begin{array}$-\frac{11}{12}\leq k\leq\frac{1}{12}\\-\frac{1}{12}\leq m\leq\frac{11}{12} \end{array}\right. \Rightarrow \left[\begin{array}$k=0: \ x_1= \frac{5 \pi }{6}+0= \frac{5 \pi }{6} \\m=0: \ x_2= -\frac{5 \pi }{6}+0= -\frac{5 \pi }{6} \end{array}\right.[/latex] Ответ: -5π/6; 5π/6