Question

Đã trả lời bởi chuyên gia

Answer 1

$2 \sin (x + \frac{π}{4}) + \sin (x - \frac{π}{4}) = \frac{3 \sqrt{2}}{2} < = > 2 \sin x . \cos \frac{π}{4} + 2 \cos x . \sin \frac{π}{4} + \sin x . \cos \frac{π}{4} - \cos x . \sin \frac{π}{4} = \frac{3 \sqrt{2}}{2} < = > \sqrt{2} \sin x + \sqrt{2} \cos x + \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x = \frac{3 \sqrt{2}}{2} < = > \frac{3 \sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x = \frac{3 \sqrt{2}}{2} < = > 3 \sin x - \cos x = 3 < = > \frac{3}{\sqrt{10}} \sin x - \frac{1}{\sqrt{10}} \cos x = \frac{3}{\sqrt{10}} < = > \sin x . \cos a - \cos x . \sin a = \frac{3 \sqrt{10}}{10} v ớ i a l à g ó c c ó : \cos a = \frac{3}{\sqrt{10}}; \sin a = \frac{1}{\sqrt{10}} < = > \sin (x - a) = \frac{3 \sqrt{10}}{10} < = > x - a = a r c \sin (\frac{3 \sqrt{10}}{10}) + k 2 π hoặc x - a = π - \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π, k \in Z < = > x = a + \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π hoặc x = π + a - \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π, k \in Z vậy S = \{a + \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π; π + a - \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π\}, k \in Z$ .

...Xem thêm

Answer 2

$2 \sin (x + \frac{π}{4}) + \sin (x - \frac{π}{4}) = \frac{3 \sqrt{2}}{2} < = > 2 \sin x . \cos \frac{π}{4} + 2 \cos x . \sin \frac{π}{4} + \sin x . \cos \frac{π}{4} - \cos x . \sin \frac{π}{4} = \frac{3 \sqrt{2}}{2} < = > \sqrt{2} \sin x + \sqrt{2} \cos x + \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x = \frac{3 \sqrt{2}}{2} < = > \frac{3 \sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x = \frac{3 \sqrt{2}}{2} < = > 3 \sin x - \cos x = 3 < = > \frac{3}{\sqrt{10}} \sin x - \frac{1}{\sqrt{10}} \cos x = \frac{3}{\sqrt{10}} < = > \sin x . \cos a - \cos x . \sin a = \frac{3 \sqrt{10}}{10} v ớ i a l à g ó c c ó : \cos a = \frac{3}{\sqrt{10}}; \sin a = \frac{1}{\sqrt{10}} < = > \sin (x - a) = \frac{3 \sqrt{10}}{10} < = > x - a = a r c \sin (\frac{3 \sqrt{10}}{10}) + k 2 π hoặc x - a = π - \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π, k \in Z < = > x = a + \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π hoặc x = π + a - \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π, k \in Z vậy S = \{a + \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π; π + a - \arcsin (\frac{3 \sqrt{10}}{10}) + k 2 π\}, k \in Z$ .

1 câu trả lời 19548

Câu hỏi hot cùng chủ đề

Bài viết mới nhất Lớp 11