Question

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Answer 1

$3 (\sin x + \cos x) + 4 \sin x . \cos x = - 3 Đ ặ t t = \sin x + \cos x = \sqrt{2} \sin (x + \frac{π}{4}), t \in [- \sqrt{2}; \sqrt{2}] = > t^{2} = 1 + 2 \sin x . \cos x = > 2 \sin x . \cos x = t^{2} - 1 T a c ó : 3 t + 2 . (t^{2} - 1) = - 3 < = > 2 t^{2} + 3 t + 1 = 0 < = > t = - 1 (t m) h o ặ c t = - \frac{1}{2} (t m) t = - 1 < = > \sqrt{2} \sin (x + \frac{π}{4}) = - 1 < = > \sin (x + \frac{π}{4}) = - \frac{\sqrt{2}}{2} < = > x + \frac{π}{4} = - \frac{π}{4} + k 2 π, hoặc x + \frac{π}{4} = π + \frac{π}{4} + k 2 π, k \in Z < = > x = - \frac{π}{2} + k 2 π hoặc x = π + k 2 π, k \in Z t = - \frac{1}{2} < = > \sqrt{2} \sin (x + \frac{π}{4}) = - \frac{1}{2} < = > \sin (x + \frac{π}{4}) = - \frac{\sqrt{2}}{4} < = > x + \frac{π}{4} = \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π hoặc x + \frac{π}{4} = π - \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π, k \in Z < = > x = - \frac{π}{4} + \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π hoặc x = \frac{3 π}{4} - \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π, k \in Z Vậy S = \{- \frac{π}{4} + \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π; \frac{3 π}{4} - \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π; - \frac{π}{2} + k 2 π; π + k 2 π\} k \in Z$ .

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Answer 2

$3 (\sin x + \cos x) + 4 \sin x . \cos x = - 3 Đ ặ t t = \sin x + \cos x = \sqrt{2} \sin (x + \frac{π}{4}), t \in [- \sqrt{2}; \sqrt{2}] = > t^{2} = 1 + 2 \sin x . \cos x = > 2 \sin x . \cos x = t^{2} - 1 T a c ó : 3 t + 2 . (t^{2} - 1) = - 3 < = > 2 t^{2} + 3 t + 1 = 0 < = > t = - 1 (t m) h o ặ c t = - \frac{1}{2} (t m) t = - 1 < = > \sqrt{2} \sin (x + \frac{π}{4}) = - 1 < = > \sin (x + \frac{π}{4}) = - \frac{\sqrt{2}}{2} < = > x + \frac{π}{4} = - \frac{π}{4} + k 2 π, hoặc x + \frac{π}{4} = π + \frac{π}{4} + k 2 π, k \in Z < = > x = - \frac{π}{2} + k 2 π hoặc x = π + k 2 π, k \in Z t = - \frac{1}{2} < = > \sqrt{2} \sin (x + \frac{π}{4}) = - \frac{1}{2} < = > \sin (x + \frac{π}{4}) = - \frac{\sqrt{2}}{4} < = > x + \frac{π}{4} = \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π hoặc x + \frac{π}{4} = π - \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π, k \in Z < = > x = - \frac{π}{4} + \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π hoặc x = \frac{3 π}{4} - \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π, k \in Z Vậy S = \{- \frac{π}{4} + \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π; \frac{3 π}{4} - \arcsin (- \frac{\sqrt{2}}{4}) + k 2 π; - \frac{π}{2} + k 2 π; π + k 2 π\} k \in Z$ .

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