вправа 23.30 гдз 11 клас алгебра Істер 2019
Вправа 23.30
Умова:
Розв'яжіть нерівність:
Розв'яжіть нерівність:
Відповідь ГДЗ:
1) cos2x ≥ 0
нехай: 2x = t, cost ≥ 0
t1 = arccos0 = π/2
t2 = arccos0 = 3π/2 \begin{equation} \frac{3\Pi }{2}+\Pi n<t<\frac{\Pi }{2}+\Pi n,n\in Z \end{equation} \begin{equation} \frac{3\Pi }{2}+\Pi n<2x<\frac{\Pi }{2}+\Pi n \end{equation} \begin{equation} \frac{3\Pi }{4}+\frac{n\Pi }{2}<x<\frac{\Pi }{4}+\frac{\Pi n}{2}; \end{equation} \begin{equation} 2)sin\frac{x}{6}<-\frac{1}{2} \end{equation} \begin{equation} \frac{x}{6}=t \end{equation} \begin{equation} t_{1}=arcsin(-\frac{1}{2})=\frac{7\Pi }{6} \end{equation} \begin{equation} t_{2}=arcsin(-\frac{1}{2})=\frac{11\Pi }{6} \end{equation} \begin{equation} \frac{7\Pi }{6}+\Pi n<t<\frac{11\Pi }{6}+\Pi n,n\in Z \end{equation} \begin{equation} 7\Pi+6\Pi n<\frac{x}{6}<11\Pi+6\Pi n,n\in Z; \end{equation} \begin{equation} 3)tg(x+\frac{\Pi }{4})>1 \end{equation} \begin{equation} x+\frac{\Pi }{4}=t,tgt>1 \end{equation} \begin{equation} t_{1}=arctg1=\frac{\Pi }{4} \end{equation} \begin{equation} t_{2}=arctg1=\frac{5}{4}\Pi \end{equation} \begin{equation} \frac{\Pi }{4}+\Pi n<t<\frac{5\Pi }{4}+n\Pi ,n\in Z \end{equation} \begin{equation} \frac{\Pi }{4}+\Pi n<x+\frac{\Pi }{4}<\frac{5\Pi }{4}+n\Pi \end{equation} \begin{equation} -\frac{\Pi }{4}+\frac{\Pi }{4}+n\Pi<x<\frac{5\Pi }{4}-\frac{\Pi }{4}+n\Pi \end{equation} \begin{equation} n\Pi<x<\Pi +n\Pi ,n\in Z; \end{equation} \begin{equation} 4)ctg(x-\frac{\Pi }{3})\leq \frac{1}{\sqrt{3}} \end{equation} \begin{equation} t=x-\frac{\Pi }{3},ctgt\leq \frac{1}{\sqrt{3}} \end{equation} \begin{equation} t_{1}=arcctg\frac{1}{\sqrt{3}}=\frac{\Pi }{3} \end{equation} \begin{equation} t_{2}=arcctg\frac{1}{\sqrt{3}}=\frac{4\Pi }{3} \end{equation} \begin{equation} \frac{\Pi }{3}+\Pi n\leq t\leq \frac{4\Pi }{3}+n\Pi ,n\in Z \end{equation} \begin{equation} \frac{\Pi }{3}+\Pi n\leq x-\frac{\Pi }{3}\leq \frac{4\Pi }{3}+n\Pi ,n\in Z \end{equation} \begin{equation} \frac{\Pi }{3}+\frac{\Pi }{3}+\Pi n\leq x\leq \frac{\Pi }{3}+\frac{\Pi }{3}+n\Pi \end{equation} \begin{equation} \frac{2\Pi }{3}+n\Pi \leq x\leq \frac{5\Pi }{3}+n\Pi ,n\in Z. \end{equation}
1) cos2x ≥ 0
нехай: 2x = t, cost ≥ 0
t1 = arccos0 = π/2
t2 = arccos0 = 3π/2 \begin{equation} \frac{3\Pi }{2}+\Pi n<t<\frac{\Pi }{2}+\Pi n,n\in Z \end{equation} \begin{equation} \frac{3\Pi }{2}+\Pi n<2x<\frac{\Pi }{2}+\Pi n \end{equation} \begin{equation} \frac{3\Pi }{4}+\frac{n\Pi }{2}<x<\frac{\Pi }{4}+\frac{\Pi n}{2}; \end{equation} \begin{equation} 2)sin\frac{x}{6}<-\frac{1}{2} \end{equation} \begin{equation} \frac{x}{6}=t \end{equation} \begin{equation} t_{1}=arcsin(-\frac{1}{2})=\frac{7\Pi }{6} \end{equation} \begin{equation} t_{2}=arcsin(-\frac{1}{2})=\frac{11\Pi }{6} \end{equation} \begin{equation} \frac{7\Pi }{6}+\Pi n<t<\frac{11\Pi }{6}+\Pi n,n\in Z \end{equation} \begin{equation} 7\Pi+6\Pi n<\frac{x}{6}<11\Pi+6\Pi n,n\in Z; \end{equation} \begin{equation} 3)tg(x+\frac{\Pi }{4})>1 \end{equation} \begin{equation} x+\frac{\Pi }{4}=t,tgt>1 \end{equation} \begin{equation} t_{1}=arctg1=\frac{\Pi }{4} \end{equation} \begin{equation} t_{2}=arctg1=\frac{5}{4}\Pi \end{equation} \begin{equation} \frac{\Pi }{4}+\Pi n<t<\frac{5\Pi }{4}+n\Pi ,n\in Z \end{equation} \begin{equation} \frac{\Pi }{4}+\Pi n<x+\frac{\Pi }{4}<\frac{5\Pi }{4}+n\Pi \end{equation} \begin{equation} -\frac{\Pi }{4}+\frac{\Pi }{4}+n\Pi<x<\frac{5\Pi }{4}-\frac{\Pi }{4}+n\Pi \end{equation} \begin{equation} n\Pi<x<\Pi +n\Pi ,n\in Z; \end{equation} \begin{equation} 4)ctg(x-\frac{\Pi }{3})\leq \frac{1}{\sqrt{3}} \end{equation} \begin{equation} t=x-\frac{\Pi }{3},ctgt\leq \frac{1}{\sqrt{3}} \end{equation} \begin{equation} t_{1}=arcctg\frac{1}{\sqrt{3}}=\frac{\Pi }{3} \end{equation} \begin{equation} t_{2}=arcctg\frac{1}{\sqrt{3}}=\frac{4\Pi }{3} \end{equation} \begin{equation} \frac{\Pi }{3}+\Pi n\leq t\leq \frac{4\Pi }{3}+n\Pi ,n\in Z \end{equation} \begin{equation} \frac{\Pi }{3}+\Pi n\leq x-\frac{\Pi }{3}\leq \frac{4\Pi }{3}+n\Pi ,n\in Z \end{equation} \begin{equation} \frac{\Pi }{3}+\frac{\Pi }{3}+\Pi n\leq x\leq \frac{\Pi }{3}+\frac{\Pi }{3}+n\Pi \end{equation} \begin{equation} \frac{2\Pi }{3}+n\Pi \leq x\leq \frac{5\Pi }{3}+n\Pi ,n\in Z. \end{equation}