Тригонометриялық теңсіздіктерді шешу (Рустюмова 4.3.1B)

Шешуі: $$\arctg 1 + \pi n \le x + \dfrac{\pi }{4} \lt \dfrac{\pi }{2} + \pi n;\quad n \in Z$$ $${\dfrac{\pi }{4} – \dfrac{\pi }{4} + \pi n \le x \lt \dfrac{\pi }{2} – \dfrac{\pi }{4} + \pi n;\quad n \in Z}$$ $${\pi n \le x \lt \dfrac{\pi }{4} + \pi n;\quad n \in Z}$$ Жауабы: ${\left[ {\pi n;\,\,\dfrac{\pi }{4} + \pi n} \right)}$

Шешуі: $${\sin \left( {x – \dfrac{\pi }{3}} \right) \le \dfrac{{\sqrt 3 }}{2}}$$ $${ – \pi – \dfrac{\pi }{3} + 2\pi n \le x – \dfrac{\pi }{3} \le \dfrac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${ – \pi – \dfrac{\pi }{3} + \dfrac{\pi }{3} + 2\pi n \le x \le \dfrac{\pi }{3} + \dfrac{\pi }{3} + 2\pi ;\quad n \in Z}$$ $${ – \pi + 2\pi n \le x \le \dfrac{{2\pi }}{3} + 2\pi ;\quad n \in Z}$$ Жауабы: ${\left[ { – \pi + 2\pi n;\dfrac{{2\pi }}{3} + 2\pi n} \right]}$

Шешуі: $$\cos 2x \ge \dfrac{1}{2}$$ $${ – \dfrac{\pi }{3} + 2\pi n \le 2x \le \dfrac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{\pi }{6} + \pi n \le x \le \dfrac{\pi }{6} + \pi n;\quad n \in Z}$$ Жауабы: ${\left[ { – \dfrac{\pi }{6} + \pi n;\,\,\dfrac{\pi }{6} + \pi n} \right]}$

Шешуі: $$\sin \left( {2x – \dfrac{\pi }{3}} \right) \ge – \dfrac{1}{2}$$ $${ – \dfrac{\pi }{6} + 2\pi n \le 2x – \dfrac{\pi }{3} \le \pi + \dfrac{\pi }{6} + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{\pi }{6} + \dfrac{\pi }{3} + 2\pi n \le 2x \le \pi + \dfrac{\pi }{6} + \dfrac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${\dfrac{\pi }{6} + 2\pi n \le 2x \le \dfrac{{3\pi }}{2} + 2\pi n;\quad n \in Z}$$ $${\dfrac{\pi }{{12}} + \pi n \le x \le \dfrac{{3\pi }}{4} + \pi n;\quad n \in Z}$$

Шешуі: $${\cos 2x \le – \dfrac{{\sqrt 3 }}{2}}$$ $${\pi – \dfrac{\pi }{6} + 2\pi n \le 2x \le \pi + \dfrac{\pi }{6} + 2\pi n;\quad n \in Z}$$ $${\dfrac{{5\pi }}{6} + 2\pi n \le 2x \le \dfrac{{7\pi }}{6} + 2\pi n;\quad n \in Z}$$ $${\dfrac{{5\pi }}{{12}} + \pi n \le x \le \dfrac{{7\pi }}{{12}} + \pi n;\quad n \in Z}$$ Жауабы: ${\left[ {\dfrac{{5\pi }}{{12}} + \pi n;\dfrac{{7\pi }}{{12}} + \pi n} \right]}$

Шешуі: $${ – \dfrac{\pi }{4} + 2\pi n \lt 2x – 1 \lt \pi + \dfrac{\pi }{4} + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{\pi }{4} + 1 + 2\pi n \lt 2x \lt \dfrac{{5\pi }}{4} + 1 + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{\pi }{8} + \dfrac{1}{2} + \pi n \lt x \lt \dfrac{{5\pi }}{8} + \dfrac{1}{2} + \pi n;\quad n \in Z}$$ Жауабы: $\left( { – \dfrac{\pi }{8} + \dfrac{1}{2} + \pi n;\dfrac{{5\pi }}{8} + \dfrac{1}{2} + \pi n} \right)$

Шешуі: $$\left\{ {\begin{array}{*{20}{l}}{\cos x \gt 0}\\{\cos x \le \dfrac{1}{2}}\end{array}} \right.$$ $${ – \dfrac{\pi }{2} + 2\pi n \lt x \le – \dfrac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${\dfrac{\pi }{3} + 2\pi n \le x \lt \dfrac{\pi }{2} + 2\pi n;\quad n \in Z}$$ Жауабы: $\left( { – \dfrac{\pi }{2} + 2\pi n;\,\, – \dfrac{\pi }{3} + 2\pi n} \right] \cup \left[ { – \dfrac{\pi }{3} + 2\pi n;\,\,\dfrac{\pi }{2} + 2\pi n} \right)$

Шешуі: $$\ctg x – 1 \ge 0$$ $$\ctg x \ge 1$$ $$\pi n \lt x \le \dfrac{\pi }{4} + \pi n;\quad n \in Z$$ Жауабы: $\left( {\pi n;\dfrac{\pi }{4} + \pi n} \right]$

Шешуі: $$2\pi n \lt x \lt \pi + 2\pi n;\quad n \in Z$$ $${\cos x = – \dfrac{1}{2}}$$ $${x = \pm \arccos \left( { – \dfrac{1}{2}} \right) + 2\pi n;\quad n \in Z}$$ $${x = \pm \dfrac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ $${x = \dfrac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ $${x = \dfrac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ $${0 \le \dfrac{{2\pi }}{3} + 2\pi n \le 4\pi }$$ $${ – \dfrac{{2\pi }}{3} \le 2\pi n \le 4\pi – \dfrac{{2\pi }}{3}}$$ $${ – \dfrac{{2\pi }}{3} \le 2\pi n \le \dfrac{{10\pi }}{3}}$$ $${ – \dfrac{2}{3} \le 2n \le \dfrac{{10}}{3}}$$ $${ – \dfrac{1}{3} \le n \le \dfrac{5}{3}}$$ $${n = 0;\,\,1}$$ $${n = 0;\quad x = \dfrac{{2\pi }}{3} + 2\pi \cdot 0 = \dfrac{{2\pi }}{3}}$$ $${n = 1;\quad x = \dfrac{{2\pi }}{3} + 2\pi \cdot 1 = \dfrac{{2\pi }}{3} + 2\pi = \dfrac{{8\pi }}{3}}$$ $${\dfrac{{2\pi }}{3} + \dfrac{{8\pi }}{3} = \dfrac{{10\pi }}{3}}$$ Жауабы: $\dfrac{10\pi}{3}$

Шешуі: $$ – \dfrac{\pi }{2} + \pi n \lt \dfrac{x}{4} \lt \pi n;\quad n \in Z$$ $$ – 2\pi + 4\pi n \lt x \lt 4\pi n;\quad n \in Z$$ Жауабы: $( – 2\pi + 4\pi n;\,\,4\pi n)$

Шешуі: \[{\left\{ {\begin{array}{*{20}{l}} {\ctg x \lt \sqrt 3 } \\ {\ctg x \gt – \sqrt 3 } \end{array}} \right.}\] $${\dfrac{\pi }{6} + \pi n \lt x \lt \pi – \dfrac{\pi }{6} + \pi n;\quad n \in Z}$$ $${\dfrac{\pi }{6} + \pi n \lt x \lt \dfrac{{5\pi }}{6} + \pi n;\quad n \in Z}$$ $${\left( {\dfrac{\pi }{6} + \pi n;\,\,\dfrac{{5\pi }}{6} + \pi n} \right)}$$

Шешуі: $${\dfrac{\pi }{4} + \pi n \le 2x \lt \dfrac{\pi }{2} + \pi n;\quad n \in Z}$$ $${\dfrac{\pi }{8} + \dfrac{{\pi n}}{2} \le x \lt \dfrac{\pi }{4} + \dfrac{{\pi n}}{2};\quad n \in Z}$$ $${\left[ {\dfrac{\pi }{8} + \dfrac{{\pi n}}{2};\,\,\dfrac{\pi }{4} + \dfrac{{\pi n}}{2}} \right)}$$

Шешуі: $$\left[ \begin{array}{l}\sin x \le \dfrac{{\sqrt 2 }}{2}\\\sin x \ge – \dfrac{{\sqrt 2 }}{2}\end{array} \right.$$ $$ – \dfrac{\pi }{4} + \pi n \le x \le \dfrac{\pi }{4} + \pi n;\quad n \in Z$$ Жауабы: $\left[ { – \dfrac{\pi }{4} + \pi n;\,\,\dfrac{\pi }{4} + \pi n} \right]$

Шешуі: $$\left\{ \begin{array}{l} \sin x \lt {\left( {\dfrac{1}{2}} \right)^1}\\ \sin x \gt 0,\quad \text{(Анықталу облысы)} \end{array} \right.$$ ${2\pi n \lt x \lt \dfrac{\pi }{6} + 2\pi n;\quad n \in Z} $ және $ \pi – \dfrac{\pi }{6} + 2\pi n \lt x \lt \pi + 2\pi n;\quad n \in Z$
Жауабы: ${\left( {2\pi n;\,\,\dfrac{\pi }{6} + 2\pi n} \right) \cup \left( {\dfrac{{5\pi }}{6} + 2\pi n;\,\,\pi + 2\pi n} \right)}$

Шешуі: $${\pi – \dfrac{\pi }{4} + \pi n \lt x + \dfrac{\pi }{3} \lt \pi + \pi n;\quad n \in Z}$$ $${\dfrac{{3\pi }}{4} – \dfrac{\pi }{3} + \pi n \lt x \lt \pi – \dfrac{\pi }{3} + \pi n;\quad n \in Z}$$ $${\dfrac{{5\pi }}{{12}} + \pi n \lt x \lt \dfrac{{2\pi }}{3} + \pi n;\quad n \in Z}$$ Жауабы: ${\left( {\dfrac{{5\pi }}{{12}} + \pi n;\,\,\dfrac{{2\pi }}{3} + \pi n} \right)}$

Шешуі: $$\left[ \begin{array}{l}\tg x \ge \sqrt 3 \\ \tg x \le – \sqrt 3 \end{array} \right.$$ $$ – \dfrac{\pi }{2} + \pi n \lt x \le – \dfrac{\pi }{3} + \pi n;\quad n \in Z{\rm{ }}$$ $${\rm{ }}\dfrac{\pi }{3} + \pi n \le x \lt \dfrac{\pi }{2} + \pi n;\quad n \in Z$$ Жауабы: ${\left( { – \dfrac{\pi }{2} + \pi n; – \dfrac{\pi }{3} + \pi n} \right] \cup \left[ {\dfrac{\pi }{3} + \pi n;\dfrac{\pi }{2} + \pi n} \right)}$

Шешуі: $${\tg^2 x -\dfrac{1}{3} \ge 0}$$ $$\left(\tg x-\dfrac{\sqrt 3}{3}\right)\left(\tg x+\dfrac{\sqrt 3}{3}\right) \ge 0$$ $\tg x=t$ жаңа айнымалысын енгізейік: $$\left( t-\dfrac{\sqrt 3}{3}\right)\left( t-\dfrac{\sqrt 3}{3}\right) \ge 0$$ $$\left[ \begin{array}{l}\tg x \le – \dfrac{{\sqrt 3 }}{3}\\ \tg x \ge \dfrac{{\sqrt 3 }}{3}\end{array} \right.$$ Жауабы: $$\left( { – \dfrac{\pi }{2} + \pi n;\,\, – \dfrac{\pi }{6} + \pi n} \right] \cup \left[ {\dfrac{\pi }{6} + \pi n;\,\,\dfrac{\pi }{2} + \pi n} \right)$$

Шешуі: $${ – \pi – \dfrac{\pi }{4} + 2\pi n \lt \dfrac{{3x}}{2} + \dfrac{\pi }{{12}} \lt \dfrac{\pi }{4} + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{{5\pi }}{4} – \dfrac{\pi }{{12}} + 2\pi n \lt \dfrac{{3x}}{2} \lt \dfrac{\pi }{4} – \dfrac{\pi }{{12}} + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{{16\pi }}{{12}} + 2\pi n \lt \dfrac{{3x}}{2} \lt \dfrac{{2\pi }}{{12}} + 2\pi n;\quad n \in Z}$$ $${ – \dfrac{{32\pi }}{{36}} + \dfrac{{4\pi n}}{3} \lt x \lt \dfrac{{4\pi }}{{36}} + \dfrac{{4\pi n}}{3};\quad n \in Z}$$ $${ – \dfrac{{8\pi }}{9} + \dfrac{{4\pi n}}{3} \lt x \lt \dfrac{\pi }{9} + \dfrac{{4\pi n}}{3};\quad n \in Z}$$ Жауабы: ${\left( { – \dfrac{{8\pi }}{9} + \dfrac{{4\pi n}}{3};\,\,\dfrac{\pi }{9} + \dfrac{{4\pi n}}{3}} \right)}$

Шешуі: $$\left[ \begin{array}{l}\cos x \ge \dfrac{{\sqrt 2 }}{2}\\\cos x \le – \dfrac{{\sqrt 2 }}{2}\end{array} \right.$$ $${ – \dfrac{\pi }{4} + \pi n \le x \le \dfrac{\pi }{4} + \pi n;\quad n \in Z}$$ $${\left[ { – \dfrac{\pi }{4} + \pi n;\,\,\dfrac{\pi }{4} + \pi n} \right]}$$

Шешуі: $$\left\{ \begin{array}{l}\sin x \le \dfrac{{\sqrt 3 }}{2}\\\sin x \ge – \dfrac{{\sqrt 3 }}{2}\end{array} \right.$$ $${ – \dfrac{\pi }{3} + \pi n \le x \le \dfrac{\pi }{3} + \pi n;\quad n \in R}$$ Жауабы: ${\left[ { – \dfrac{\pi }{3} + \pi n;\,\,\dfrac{\pi }{3} + \pi n} \right]}$