Question

Đã trả lời bởi chuyên gia

Answer 1

$a, \sin 3 x + \cos 3 x = \sqrt{2} \sin 2 x < = > \sqrt{2} \sin (3 x + \frac{π}{4}) = \sqrt{2} \sin 2 x < = > \sin (3 x + \frac{π}{4}) = \sin 2 x < = > 3 x + \frac{π}{4} = 2 x + k 2 π hoặc 3 x + \frac{π}{4} = π - 2 x + k 2 π, k \in Z < = > x = - \frac{π}{4} + k 2 π hoặc 5 x = \frac{3 π}{4} + k 2 π, k \in Z < = > x = - \frac{π}{4} + k 2 π hoặc x = \frac{3 π}{20} + \frac{k 2 π}{5}, k \in Z Vậy S = \{- \frac{π}{4} + k 2 π; \frac{3 π}{20} + \frac{k 2 π}{5}\}, k \in Z$

$b, (\cos x + \sqrt{2} \sin x) (1 - \cos x) = \sin^{2} x < = > (\cos x + \sqrt{2} \sin x) (1 - \cos x) = 1 - \cos^{2} x < = > (\cos x + \sqrt{2} \sin x) (1 - \cos x) = (1 - \cos x) (1 + \cos x) < = > (\cos x + \sqrt{2} \sin x) (1 - \cos x) - (1 - \cos x) (1 + \cos x) = 0 < = > (1 - \cos x) (\cos x + \sqrt{2} \sin x - 1 - \cos x) = 0 < = > (1 - \cos x) (\sqrt{2} \sin x - 1) = 0 < = > 1 - \cos x = 0 h o ặ c \sqrt{2} \sin x - 1 = 0 < = > \cos x = 1 h o ặ c \sin x = \frac{1}{\sqrt{2}} < = > x = k 2 π hoặc x = \frac{π}{4} + k 2 π hoặc x = \frac{3 π}{4} + k 2 π, k \in Z Vậy S = \{k 2 π; \frac{π}{4} + k 2 π; \frac{3 π}{4} + k 2 π\}, k \in Z$ ..

...Xem thêm

Answer 2

$a, \sin 3 x + \cos 3 x = \sqrt{2} \sin 2 x < = > \sqrt{2} \sin (3 x + \frac{π}{4}) = \sqrt{2} \sin 2 x < = > \sin (3 x + \frac{π}{4}) = \sin 2 x < = > 3 x + \frac{π}{4} = 2 x + k 2 π hoặc 3 x + \frac{π}{4} = π - 2 x + k 2 π, k \in Z < = > x = - \frac{π}{4} + k 2 π hoặc 5 x = \frac{3 π}{4} + k 2 π, k \in Z < = > x = - \frac{π}{4} + k 2 π hoặc x = \frac{3 π}{20} + \frac{k 2 π}{5}, k \in Z Vậy S = \{- \frac{π}{4} + k 2 π; \frac{3 π}{20} + \frac{k 2 π}{5}\}, k \in Z$

$b, (\cos x + \sqrt{2} \sin x) (1 - \cos x) = \sin^{2} x < = > (\cos x + \sqrt{2} \sin x) (1 - \cos x) = 1 - \cos^{2} x < = > (\cos x + \sqrt{2} \sin x) (1 - \cos x) = (1 - \cos x) (1 + \cos x) < = > (\cos x + \sqrt{2} \sin x) (1 - \cos x) - (1 - \cos x) (1 + \cos x) = 0 < = > (1 - \cos x) (\cos x + \sqrt{2} \sin x - 1 - \cos x) = 0 < = > (1 - \cos x) (\sqrt{2} \sin x - 1) = 0 < = > 1 - \cos x = 0 h o ặ c \sqrt{2} \sin x - 1 = 0 < = > \cos x = 1 h o ặ c \sin x = \frac{1}{\sqrt{2}} < = > x = k 2 π hoặc x = \frac{π}{4} + k 2 π hoặc x = \frac{3 π}{4} + k 2 π, k \in Z Vậy S = \{k 2 π; \frac{π}{4} + k 2 π; \frac{3 π}{4} + k 2 π\}, k \in Z$ ..

1 câu trả lời 314

Câu hỏi hot cùng chủ đề

Bài viết mới nhất Lớp 11