Question

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Answer 1

$a, \cos x + 5 \sin \frac{x}{2} - 3 = 0 < = > 1 - 2 \sin^{2} \frac{x}{2} + 5 \sin x - 3 = 0 < = > - 2 \sin^{2} \frac{x}{2} + 5 \sin x - 2 = 0 < = > \sin \frac{x}{2} = 2 (l o ạ i) h o ặ c \sin \frac{x}{2} = \frac{1}{2} < = > \frac{x}{2} = \frac{π}{6} + k 2 π hoặc \frac{x}{2} = π - \frac{π}{6} + k 2 π, k \in Z < = > x = \frac{π}{3} + k 4 π hoặc x = \frac{5 π}{3} + k 4 π, k \in Z Vậy S = \{\frac{π}{3} + k 4 π; \frac{5 π}{3} + k 4 π\}, k \in Z$

$b, \cos 4 x - \sin 2 x - 1 = 0 < = > 1 - 2 \sin^{2} 2 x - \sin 2 x - 1 = 0 < = > 2 \sin^{2} 2 x + \sin 2 x = 0 < = > \sin 2 x = 0 h o ặ c \sin 2 x = \frac{- 1}{2} < = > 2 x = k π hoặc 2 x = \frac{- π}{6} + k 2 π hoặc 2 x = π + \frac{π}{6} + k 2 π, k \in Z < = > x = \frac{kπ}{2} hoặc x = \frac{- π}{12} + kπ hoặc x = \frac{7 π}{12} + kπ, k \in Z Vậy S = \{\frac{kπ}{2}; \frac{- π}{12} + kπ; \frac{7 π}{12} + kπ\}, k \in Z$

$c, \cos 6 x - 3 \cos 3 x - 1 = 0 < = > 2 \cos^{2} 3 x - 1 - 3 \cos 3 x - 1 = 0 < = > 2 \cos^{2} 3 x - 3 \cos 3 x - 2 = 0 < = > \cos 3 x = 2 (l o ạ i) h o ặ c \cos 3 x = \frac{- 1}{2} < = > 3 x = \frac{\pm 2 π}{3} + k 2 π, k \in Z < = > x = \frac{\pm 2 π}{9} + \frac{k 2 π}{3}, k \in Z Vậy S = \{\pm \frac{2 π}{9} + \frac{k 2 π}{3}\}, k \in Z$ .

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Answer 2

$a, \cos x + 5 \sin \frac{x}{2} - 3 = 0 < = > 1 - 2 \sin^{2} \frac{x}{2} + 5 \sin x - 3 = 0 < = > - 2 \sin^{2} \frac{x}{2} + 5 \sin x - 2 = 0 < = > \sin \frac{x}{2} = 2 (l o ạ i) h o ặ c \sin \frac{x}{2} = \frac{1}{2} < = > \frac{x}{2} = \frac{π}{6} + k 2 π hoặc \frac{x}{2} = π - \frac{π}{6} + k 2 π, k \in Z < = > x = \frac{π}{3} + k 4 π hoặc x = \frac{5 π}{3} + k 4 π, k \in Z Vậy S = \{\frac{π}{3} + k 4 π; \frac{5 π}{3} + k 4 π\}, k \in Z$

$b, \cos 4 x - \sin 2 x - 1 = 0 < = > 1 - 2 \sin^{2} 2 x - \sin 2 x - 1 = 0 < = > 2 \sin^{2} 2 x + \sin 2 x = 0 < = > \sin 2 x = 0 h o ặ c \sin 2 x = \frac{- 1}{2} < = > 2 x = k π hoặc 2 x = \frac{- π}{6} + k 2 π hoặc 2 x = π + \frac{π}{6} + k 2 π, k \in Z < = > x = \frac{kπ}{2} hoặc x = \frac{- π}{12} + kπ hoặc x = \frac{7 π}{12} + kπ, k \in Z Vậy S = \{\frac{kπ}{2}; \frac{- π}{12} + kπ; \frac{7 π}{12} + kπ\}, k \in Z$

$c, \cos 6 x - 3 \cos 3 x - 1 = 0 < = > 2 \cos^{2} 3 x - 1 - 3 \cos 3 x - 1 = 0 < = > 2 \cos^{2} 3 x - 3 \cos 3 x - 2 = 0 < = > \cos 3 x = 2 (l o ạ i) h o ặ c \cos 3 x = \frac{- 1}{2} < = > 3 x = \frac{\pm 2 π}{3} + k 2 π, k \in Z < = > x = \frac{\pm 2 π}{9} + \frac{k 2 π}{3}, k \in Z Vậy S = \{\pm \frac{2 π}{9} + \frac{k 2 π}{3}\}, k \in Z$ .

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