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Answer 1

$\sin 4 x + 2 = \cos 3 x + 4 \sin x + \cos x < = > 2 \sin 2 x . \cos 2 x + 2 = (\cos 3 x + \cos x) + 4 \sin x < = > 4 \sin x . \cos x . \cos 2 x + 2 = 2 \cos 2 x . \cos x + 4 \sin x < = > 4 \sin x . \cos x . \cos 2 x - 2 \cos 2 x . \cos x + 2 - 4 \sin x = 0 < = > 2 \cos x . \cos 2 x . (2 \sin x - 1) - 2 . (2 \sin x - 1) = 0 < = > (2 \sin x - 1) . (2 \cos x . \cos 2 x - 2) = 0 < = > 2 \sin x - 1 = 0 (1) h o ặ c 2 \cos x . \cos 2 x - 2 = 0 (2)$

$(1) < = > \sin x = \frac{1}{2} < = > x = \frac{π}{6} + k 2 π hoặc x = π - \frac{π}{6} + k 2 π (k \in Z) < = > x = \frac{π}{6} + k 2 π hoặc x = \frac{5 π}{6} + k 2 π (k \in Z) (2) < = > \cos 2 x . cosx - 1 = 0 < = > (2 \cos^{2} x - 1) . cosx - 1 = 0 < = > 2 \cos^{3} x - cosx - 1 = 0 < = > cosx = 1 < = > x = k 2 π (k \in Z) Vậy S = \{\frac{π}{6} + k 2 π; \frac{5 π}{6} + k 2 π; k 2 π\}, k \in Z$ .

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Answer 2

$\sin 4 x + 2 = \cos 3 x + 4 \sin x + \cos x < = > 2 \sin 2 x . \cos 2 x + 2 = (\cos 3 x + \cos x) + 4 \sin x < = > 4 \sin x . \cos x . \cos 2 x + 2 = 2 \cos 2 x . \cos x + 4 \sin x < = > 4 \sin x . \cos x . \cos 2 x - 2 \cos 2 x . \cos x + 2 - 4 \sin x = 0 < = > 2 \cos x . \cos 2 x . (2 \sin x - 1) - 2 . (2 \sin x - 1) = 0 < = > (2 \sin x - 1) . (2 \cos x . \cos 2 x - 2) = 0 < = > 2 \sin x - 1 = 0 (1) h o ặ c 2 \cos x . \cos 2 x - 2 = 0 (2)$

$(1) < = > \sin x = \frac{1}{2} < = > x = \frac{π}{6} + k 2 π hoặc x = π - \frac{π}{6} + k 2 π (k \in Z) < = > x = \frac{π}{6} + k 2 π hoặc x = \frac{5 π}{6} + k 2 π (k \in Z) (2) < = > \cos 2 x . cosx - 1 = 0 < = > (2 \cos^{2} x - 1) . cosx - 1 = 0 < = > 2 \cos^{3} x - cosx - 1 = 0 < = > cosx = 1 < = > x = k 2 π (k \in Z) Vậy S = \{\frac{π}{6} + k 2 π; \frac{5 π}{6} + k 2 π; k 2 π\}, k \in Z$ .

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