Question

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Answer 1

$\sin^{4} x + \cos^{4} x - \cos^{2} x + \frac{1}{4} \sin^{2} x - 1 = 0 < = > {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x . \cos^{2} x - \cos^{2} x + \frac{1}{4} \sin^{2} x - 1 = 0 < = > 1 - \frac{1}{2} \sin^{2} 2 x - \frac{1}{2} - \frac{1}{2} \cos 2 x + \frac{1}{4} . (\frac{1}{2} - \frac{1}{2} \cos 2 x) - 1 = 0 < = > - \frac{1}{2} + \frac{1}{2} \cos^{2} 2 x - \frac{1}{2} - \frac{1}{2} \cos 2 x + \frac{1}{8} - \frac{1}{8} \cos 2 x = 0 < = > \frac{1}{2} \cos^{2} 2 x - \frac{5}{8} \cos 2 x - \frac{7}{8} = 0 < = > 4 \cos^{2} 2 x - 5 \cos 2 x - 7 = 0 Đ ặ t t = \cos 2 x, t \in [- 1; 1] C ó : 4 t^{2} - 5 t - 7 = 0 < = > t = \frac{5 + \sqrt{137}}{8} (l o ạ i) h o ặ c t = \frac{5 - \sqrt{137}}{8} (t m) t = \frac{5 - \sqrt{137}}{8} < = > \cos 2 x = \frac{5 - \sqrt{137}}{8} < = > 2 x = \pm a r c \cos (\frac{5 - \sqrt{137}}{8}) + k 2 π, k \in Z < = > x = \pm \frac{1}{2} \arccos (\frac{5 - \sqrt{137}}{8}) + kπ, k \in Z Vậy S = \{\pm \frac{1}{2} \arccos (\frac{5 - \sqrt{137}}{8}) + kπ\}, k \in Z$ .

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Answer 2

$\sin^{4} x + \cos^{4} x - \cos^{2} x + \frac{1}{4} \sin^{2} x - 1 = 0 < = > {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x . \cos^{2} x - \cos^{2} x + \frac{1}{4} \sin^{2} x - 1 = 0 < = > 1 - \frac{1}{2} \sin^{2} 2 x - \frac{1}{2} - \frac{1}{2} \cos 2 x + \frac{1}{4} . (\frac{1}{2} - \frac{1}{2} \cos 2 x) - 1 = 0 < = > - \frac{1}{2} + \frac{1}{2} \cos^{2} 2 x - \frac{1}{2} - \frac{1}{2} \cos 2 x + \frac{1}{8} - \frac{1}{8} \cos 2 x = 0 < = > \frac{1}{2} \cos^{2} 2 x - \frac{5}{8} \cos 2 x - \frac{7}{8} = 0 < = > 4 \cos^{2} 2 x - 5 \cos 2 x - 7 = 0 Đ ặ t t = \cos 2 x, t \in [- 1; 1] C ó : 4 t^{2} - 5 t - 7 = 0 < = > t = \frac{5 + \sqrt{137}}{8} (l o ạ i) h o ặ c t = \frac{5 - \sqrt{137}}{8} (t m) t = \frac{5 - \sqrt{137}}{8} < = > \cos 2 x = \frac{5 - \sqrt{137}}{8} < = > 2 x = \pm a r c \cos (\frac{5 - \sqrt{137}}{8}) + k 2 π, k \in Z < = > x = \pm \frac{1}{2} \arccos (\frac{5 - \sqrt{137}}{8}) + kπ, k \in Z Vậy S = \{\pm \frac{1}{2} \arccos (\frac{5 - \sqrt{137}}{8}) + kπ\}, k \in Z$ .

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