Question

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Answer 1

$1 + 2 \sin 2 x = 6 \cos^{2} x < = > 1 + 2 \sin 2 x = 3 + 3 \cos 2 x < = > 2 \sin 2 x - 3 \cos 2 x = 2 < = > \frac{2}{\sqrt{13}} \sin 2 x - \frac{3}{\sqrt{13}} \cos 2 x = \frac{2}{\sqrt{13}} < = > \sin 2 x . \cos a - \cos 2 x . \sin a = \frac{2 \sqrt{13}}{3} T r o n g đ ó : a l à g ó c c ó \cos a = \frac{2}{\sqrt{13}}; \sin a = \frac{3}{\sqrt{13}} < = > \sin (2 x - a) = \frac{2 \sqrt{13}}{3} < = > 2 x - a = a r c \sin (\frac{2 \sqrt{13}}{13}) + k 2 π hoặc 2 x - a = π - \arcsin (\frac{2 \sqrt{13}}{13}) + k 2 π, k \in Z < = > x = \frac{1}{2} a + \frac{1}{2} \arcsin (\frac{2 \sqrt{13}}{13}) + kπ hoặc x = \frac{π}{2} + \frac{1}{2} a - \arcsin (\frac{2 \sqrt{13}}{13}) + kπ, k \in Z Vậy S = \{\frac{1}{2} a + \frac{1}{2} \arcsin (\frac{2 \sqrt{13}}{13}) + kπ; \frac{π}{2} + \frac{1}{2} a - \arcsin (\frac{2 \sqrt{13}}{13}) + kπ\}, k \in Z$ .

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Answer 2

$1 + 2 \sin 2 x = 6 \cos^{2} x < = > 1 + 2 \sin 2 x = 3 + 3 \cos 2 x < = > 2 \sin 2 x - 3 \cos 2 x = 2 < = > \frac{2}{\sqrt{13}} \sin 2 x - \frac{3}{\sqrt{13}} \cos 2 x = \frac{2}{\sqrt{13}} < = > \sin 2 x . \cos a - \cos 2 x . \sin a = \frac{2 \sqrt{13}}{3} T r o n g đ ó : a l à g ó c c ó \cos a = \frac{2}{\sqrt{13}}; \sin a = \frac{3}{\sqrt{13}} < = > \sin (2 x - a) = \frac{2 \sqrt{13}}{3} < = > 2 x - a = a r c \sin (\frac{2 \sqrt{13}}{13}) + k 2 π hoặc 2 x - a = π - \arcsin (\frac{2 \sqrt{13}}{13}) + k 2 π, k \in Z < = > x = \frac{1}{2} a + \frac{1}{2} \arcsin (\frac{2 \sqrt{13}}{13}) + kπ hoặc x = \frac{π}{2} + \frac{1}{2} a - \arcsin (\frac{2 \sqrt{13}}{13}) + kπ, k \in Z Vậy S = \{\frac{1}{2} a + \frac{1}{2} \arcsin (\frac{2 \sqrt{13}}{13}) + kπ; \frac{π}{2} + \frac{1}{2} a - \arcsin (\frac{2 \sqrt{13}}{13}) + kπ\}, k \in Z$ .

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