Question

Đã trả lời bởi chuyên gia

Answer 1

$1, \tan (3 x + 3) = 5, đ k : x \neq - 1 + \frac{π}{6} + \frac{k π}{3}, k \in Z < = > 3 x + 3 = a r c \tan 5 + k π, k \in Z < = > 3 x = - 3 + \arctan 5 + kπ, k \in Z < = > x = - 1 + \arctan 5 + kπ, k \in Z (tm) Vậy S = \{- 1 + \arctan 5 + kπ\}, k \in Z 2, \tan (\frac{3 π}{5} - x) = \tan (\frac{π}{3} - 2 x), đk : x \neq \{\frac{π}{10} + kπ; - \frac{π}{12} + \frac{kπ}{2}\}, k \in Z < = > \frac{3 π}{5} - x = \frac{π}{3} - 2 x + kπ, k \in Z < = > x = - \frac{4 π}{15} + kπ, k \in Z (tm) Vậy S = \{- \frac{4 π}{15} + kπ\}, k \in Z 3, \cot (2 x - 1) = 1, đk : x \neq \frac{1}{2} + \frac{kπ}{2}, k \in Z < = > 2 x - 1 = \frac{π}{4} + kπ, k \in Z < = > 2 x = 1 + \frac{π}{4} + kπ, k \in Z < = > x = \frac{1}{2} + \frac{π}{8} + \frac{kπ}{2}, k \in Z (tm) vậy S = \{\frac{1}{2} + \frac{π}{8} + \frac{kπ}{2}\}, k \in Z$

$4, c o t (\frac{π}{4} - x) = c o t (\frac{π}{3} - 2 x), d k : x \neq \{\frac{π}{4} + k π; \frac{π}{6} + \frac{kπ}{2}\}, k \in Z < = > \frac{π}{4} - x = \frac{π}{3} - 2 x + k π, k \in Z < = > x = \frac{π}{12} + kπ, k \in Z (tm) Vậy S = \{\frac{π}{12} + kπ\}, k \in Z$ .

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Answer 2

$1, \tan (3 x + 3) = 5, đ k : x \neq - 1 + \frac{π}{6} + \frac{k π}{3}, k \in Z < = > 3 x + 3 = a r c \tan 5 + k π, k \in Z < = > 3 x = - 3 + \arctan 5 + kπ, k \in Z < = > x = - 1 + \arctan 5 + kπ, k \in Z (tm) Vậy S = \{- 1 + \arctan 5 + kπ\}, k \in Z 2, \tan (\frac{3 π}{5} - x) = \tan (\frac{π}{3} - 2 x), đk : x \neq \{\frac{π}{10} + kπ; - \frac{π}{12} + \frac{kπ}{2}\}, k \in Z < = > \frac{3 π}{5} - x = \frac{π}{3} - 2 x + kπ, k \in Z < = > x = - \frac{4 π}{15} + kπ, k \in Z (tm) Vậy S = \{- \frac{4 π}{15} + kπ\}, k \in Z 3, \cot (2 x - 1) = 1, đk : x \neq \frac{1}{2} + \frac{kπ}{2}, k \in Z < = > 2 x - 1 = \frac{π}{4} + kπ, k \in Z < = > 2 x = 1 + \frac{π}{4} + kπ, k \in Z < = > x = \frac{1}{2} + \frac{π}{8} + \frac{kπ}{2}, k \in Z (tm) vậy S = \{\frac{1}{2} + \frac{π}{8} + \frac{kπ}{2}\}, k \in Z$

$4, c o t (\frac{π}{4} - x) = c o t (\frac{π}{3} - 2 x), d k : x \neq \{\frac{π}{4} + k π; \frac{π}{6} + \frac{kπ}{2}\}, k \in Z < = > \frac{π}{4} - x = \frac{π}{3} - 2 x + k π, k \in Z < = > x = \frac{π}{12} + kπ, k \in Z (tm) Vậy S = \{\frac{π}{12} + kπ\}, k \in Z$ .

Answer 3

Cj ơi làm luôn câu 4 e vs ạ 🥺

Answer 4

Dạ đc rr e cảm ơn cj ạ 😚🥰

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