Question

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Answer 1

$(1 + \sin^{2} x) \cos x + (1 + \cos^{2} x) \sin x = 1 + \sin 2 x < = > \cos x + \sin^{2} x \cos x + \sin x + \cos^{2} x . \sin x = {(\sin x + \cos x)}^{2} < = > (\sin x + \cos x) + \sin^{2} x . \cos x + \cos^{2} x . \sin x = {(\sin x + \cos x)}^{2} < = > \sin x + \cos x + \sin x . \cos x . (\sin x + \cos x) = {(\sin x + \cos x)}^{2} Đ ặ t t = \sin x + \cos x = \sqrt{2} . \sin (x + \frac{π}{4}), t \in [- \sqrt{2}; \sqrt{2}] = > t^{2} = 1 + 2 \sin x \cos x = > \sin x \cos x = \frac{t^{2}}{2} - \frac{1}{2}$

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

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Answer 2

$(1 + \sin^{2} x) \cos x + (1 + \cos^{2} x) \sin x = 1 + \sin 2 x < = > \cos x + \sin^{2} x \cos x + \sin x + \cos^{2} x . \sin x = {(\sin x + \cos x)}^{2} < = > (\sin x + \cos x) + \sin^{2} x . \cos x + \cos^{2} x . \sin x = {(\sin x + \cos x)}^{2} < = > \sin x + \cos x + \sin x . \cos x . (\sin x + \cos x) = {(\sin x + \cos x)}^{2} Đ ặ t t = \sin x + \cos x = \sqrt{2} . \sin (x + \frac{π}{4}), t \in [- \sqrt{2}; \sqrt{2}] = > t^{2} = 1 + 2 \sin x \cos x = > \sin x \cos x = \frac{t^{2}}{2} - \frac{1}{2}$

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

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