Question

Đã trả lời bởi chuyên gia

Answer 1

$\cos^{2} x - 4 \sin x . \cos x = 0 < = > \frac{1}{2} + \frac{1}{2} \cos 2 x - 2 \sin 2 x = 0 < = > \cos 2 x - 4 \sin 2 x + 1 = 0 < = > \cos 2 x - 4 \sin 2 x = - 1 < = > \frac{1}{\sqrt{17}} \cos 2 x - \frac{4}{\sqrt{17}} \sin 2 x = \frac{- 1}{\sqrt{17}} < = > \sin a . \cos 2 x - \cos a . \sin 2 x = \frac{- \sqrt{17}}{17} T r o n g đ ó : a l à g ó c c ó \sin a = \frac{1}{\sqrt{17}}; \cos a = \frac{4}{\sqrt{17}} < = > \sin (a - 2 x) = \frac{- \sqrt{17}}{17} < = > a - 2 x = a r c \sin (- \frac{\sqrt{17}}{17}) + k 2 π hoặc a - 2 x = π - \arcsin (- \frac{\sqrt{17}}{17}) + k 2 π, k \in Z < = > x = \frac{1}{2} a - \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ hoặc x = \frac{- π}{2} + \frac{1}{2} a + \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ Vậy S = \{\frac{- π}{2} + \frac{1}{2} a + \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ; \frac{1}{2} a - \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ\}, k \in Z$ .

...Xem thêm

Answer 2

$\cos^{2} x - 4 \sin x . \cos x = 0 < = > \frac{1}{2} + \frac{1}{2} \cos 2 x - 2 \sin 2 x = 0 < = > \cos 2 x - 4 \sin 2 x + 1 = 0 < = > \cos 2 x - 4 \sin 2 x = - 1 < = > \frac{1}{\sqrt{17}} \cos 2 x - \frac{4}{\sqrt{17}} \sin 2 x = \frac{- 1}{\sqrt{17}} < = > \sin a . \cos 2 x - \cos a . \sin 2 x = \frac{- \sqrt{17}}{17} T r o n g đ ó : a l à g ó c c ó \sin a = \frac{1}{\sqrt{17}}; \cos a = \frac{4}{\sqrt{17}} < = > \sin (a - 2 x) = \frac{- \sqrt{17}}{17} < = > a - 2 x = a r c \sin (- \frac{\sqrt{17}}{17}) + k 2 π hoặc a - 2 x = π - \arcsin (- \frac{\sqrt{17}}{17}) + k 2 π, k \in Z < = > x = \frac{1}{2} a - \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ hoặc x = \frac{- π}{2} + \frac{1}{2} a + \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ Vậy S = \{\frac{- π}{2} + \frac{1}{2} a + \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ; \frac{1}{2} a - \frac{1}{2} \arcsin (- \frac{\sqrt{17}}{17}) + kπ\}, k \in Z$ .

1 câu trả lời 463

Câu hỏi hot cùng chủ đề

Bài viết mới nhất Lớp 11