Question

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Answer 1

$a, 2 \cos 10 x + \cos 5 x - 1 = 0 < = > 4 \cos^{2} 5 x - 2 + \cos 5 x - 1 = 0 < = > 4 \cos^{2} 5 x + \cos 5 x - 3 = 0 Đ ặ t t = \cos 5 x, t \in [- 1; 1] = > 4 t^{2} + t - 3 = 0 G i ả i r a đ ư ợ c t = - 1 (t m) h o ặ c t = \frac{3}{4} (t m) t = - 1 < = > \cos 5 x = - 1 < = > 5 x = π + k 2 π, k \in Z < = > x = \frac{π}{5} + \frac{k 2 π}{5}, k \in Z t = \frac{3}{4} < = > \cos 5 x = \frac{3}{4} < = > 5 x = \pm \arccos (\frac{3}{4}) + k 2 π, k \in Z < = > x = \pm \frac{1}{5} \arccos (\frac{3}{4}) + \frac{k 2 π}{5}, k \in Z Vậy S = \{\frac{π}{5} + \frac{k 2 π}{5}; \pm \frac{1}{5} \arccos (\frac{3}{4}) + \frac{k 2 π}{5}\}, k \in Z$

$b, 2 \cos^{2} x - \sin x + 1 = 0 < = > 2 - 2 \sin^{2} x - \sin x + 1 = 0 < = > - 2 \sin^{2} x - \sin x + 1 = 0 Đ ặ t t = \sin x, t \in [- 1; 1] = > - 2 t^{2} - t + 1 = 0 < = > t = - 1 (t m) h o ặ c t = \frac{1}{2} (t m) t = - 1 < = > \sin x = - 1 < = > x = - \frac{π}{2} + k 2 π, k \in Z t = \frac{1}{2} < = > sinx = \frac{1}{2} < = > x = \frac{π}{6} + k 2 π hoặc x = \frac{5 π}{6} + k 2 π, k \in Z Vậy S = \{\frac{π}{6} + k 2 π; \frac{5 π}{6} + k 2 π; \frac{- π}{2} + k 2 π\}, k \in Z$ .

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Answer 2

$a, 2 \cos 10 x + \cos 5 x - 1 = 0 < = > 4 \cos^{2} 5 x - 2 + \cos 5 x - 1 = 0 < = > 4 \cos^{2} 5 x + \cos 5 x - 3 = 0 Đ ặ t t = \cos 5 x, t \in [- 1; 1] = > 4 t^{2} + t - 3 = 0 G i ả i r a đ ư ợ c t = - 1 (t m) h o ặ c t = \frac{3}{4} (t m) t = - 1 < = > \cos 5 x = - 1 < = > 5 x = π + k 2 π, k \in Z < = > x = \frac{π}{5} + \frac{k 2 π}{5}, k \in Z t = \frac{3}{4} < = > \cos 5 x = \frac{3}{4} < = > 5 x = \pm \arccos (\frac{3}{4}) + k 2 π, k \in Z < = > x = \pm \frac{1}{5} \arccos (\frac{3}{4}) + \frac{k 2 π}{5}, k \in Z Vậy S = \{\frac{π}{5} + \frac{k 2 π}{5}; \pm \frac{1}{5} \arccos (\frac{3}{4}) + \frac{k 2 π}{5}\}, k \in Z$

$b, 2 \cos^{2} x - \sin x + 1 = 0 < = > 2 - 2 \sin^{2} x - \sin x + 1 = 0 < = > - 2 \sin^{2} x - \sin x + 1 = 0 Đ ặ t t = \sin x, t \in [- 1; 1] = > - 2 t^{2} - t + 1 = 0 < = > t = - 1 (t m) h o ặ c t = \frac{1}{2} (t m) t = - 1 < = > \sin x = - 1 < = > x = - \frac{π}{2} + k 2 π, k \in Z t = \frac{1}{2} < = > sinx = \frac{1}{2} < = > x = \frac{π}{6} + k 2 π hoặc x = \frac{5 π}{6} + k 2 π, k \in Z Vậy S = \{\frac{π}{6} + k 2 π; \frac{5 π}{6} + k 2 π; \frac{- π}{2} + k 2 π\}, k \in Z$ .

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