Question

Đã trả lời bởi chuyên gia

Answer 1

HD:

$4 \sin^{2} x + \sin x + \sqrt{3} \cos x = 2 \Leftrightarrow \sin x + \sqrt{3} \cos x = 2 - 4 \sin^{2} x \Leftrightarrow \sin x + \sqrt{3} \cos x = 2 \cos 2 x \Leftrightarrow \frac{\sin x + \sqrt{3} \cos x}{2} = \cos 2 x \Leftrightarrow \cos x . \cos \frac{π}{6} + \sin x \sin \frac{π}{6} = \cos 2 x \Leftrightarrow \cos (x - \frac{π}{6}) = \cos 2 x \Leftrightarrow x - \frac{π}{6} = 2 x + k 2 π h o ặ c x - \frac{π}{6} = - 2 x + k 2 π, k \in Z \Leftrightarrow x = \frac{- π}{6} + k 2 π h o ặ c 3 x = \frac{π}{6} + k 2 π, k \in Z \Leftrightarrow x = \frac{- π}{6} + k 2 π h o ặ c x = \frac{π}{18} + \frac{k 2 π}{3}, k \in Z V ậ y p t đ ã c h o c ó n g h i ệ m l à x = \frac{- π}{6} + k 2 π v à x = \frac{π}{18} + \frac{k 2 π}{3}, k \in Z$

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Answer 2

HD:

$4 \sin^{2} x + \sin x + \sqrt{3} \cos x = 2 \Leftrightarrow \sin x + \sqrt{3} \cos x = 2 - 4 \sin^{2} x \Leftrightarrow \sin x + \sqrt{3} \cos x = 2 \cos 2 x \Leftrightarrow \frac{\sin x + \sqrt{3} \cos x}{2} = \cos 2 x \Leftrightarrow \cos x . \cos \frac{π}{6} + \sin x \sin \frac{π}{6} = \cos 2 x \Leftrightarrow \cos (x - \frac{π}{6}) = \cos 2 x \Leftrightarrow x - \frac{π}{6} = 2 x + k 2 π h o ặ c x - \frac{π}{6} = - 2 x + k 2 π, k \in Z \Leftrightarrow x = \frac{- π}{6} + k 2 π h o ặ c 3 x = \frac{π}{6} + k 2 π, k \in Z \Leftrightarrow x = \frac{- π}{6} + k 2 π h o ặ c x = \frac{π}{18} + \frac{k 2 π}{3}, k \in Z V ậ y p t đ ã c h o c ó n g h i ệ m l à x = \frac{- π}{6} + k 2 π v à x = \frac{π}{18} + \frac{k 2 π}{3}, k \in Z$

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