giải phương trình sau: 1. $\frac{1+sin2x+cos2x}{1+cot^{2}x }$ = $\sqrt[]{2}$ sinx.sin2x 2. 2. $\frac{2.(cos^{6}x+sin^{6}x-sinx.cosx)}{\sqrt[]{2}- 2s

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giải phương trình sau:
1. $\frac{1+sin2x+cos2x}{1+cot^{2}x }$ = $\sqrt[]{2}$ sinx.sin2x
2. 2. $\frac{2.(cos^{6}x+sin^{6}x-sinx.cosx)}{\sqrt[]{2}- 2sinx}$ =0

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Alice 1 năm 2021-09-25T15:00:15+00:00 1 Answers 12 views 0

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    2021-09-25T15:01:30+00:00

    1. DK: $\sin x \neq 0$ hay $x \neq k\pi$.

    Nhân chéo lên ta có

    $1 + \sin(2x) + \cos(2x) = \sqrt{2} \sin x .2\sin x \cos x (1 + \dfrac{\cos^2x}{\sin^2x})$

    $<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} .\sin^2x \cos x(1 + \dfrac{\cos^2x}{\sin^2x})$

    $<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} .\sin^2x \cos x + 2\sqrt{2} \cos^3x$

    $<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} .\cos x(\sin^2x + \cos^2x)$

    $<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} \cos x$

    $<-> 1 + 2\sin x \cos x + 2\cos^2x – 1 = 2\sqrt{2} \cos x$

    $<-> \sin x \cos x + \cos^2x = \sqrt{2} \cos x$

    Vậy $\cos x = 0$ nên $x = \dfrac{\pi}{2} + k\pi$ hoặc

    $\sin x + \cos x = \sqrt{2}$

    $<-> \sqrt{2} (\dfrac{1}{\sqrt{2}} \sin x + \dfrac{1}{\sqrt{2}} \cos x ) = \sqrt{2}$

    $<-> \sin x \cos \dfrac{\pi}{4} + \sin \dfrac{\pi}{4} \cos x = 1$

    $<-> \sin(x + \dfrac{\pi}{4}) = 1$

    Vậy $x + \dfrac{\pi}{4} = \dfrac{\pi}{2} + 2k\pi$ hay $x = \dfrac{\pi}{4} + 2k\pi$.

    Vậy nghiệm của ptrinh là $x = \dfrac{\pi}{2} + k\pi$ hoặc $x = \dfrac{\pi}{4} + 2k\pi$.

    2. ĐK: $\sin x \neq \dfrac{\sqrt{2}}{2}$ hay $x \neq \dfrac{\pi}{4} + 2k\pi$ hoặc $x \neq \dfrac{3\pi}{4} + 2k\pi$.

    Ptrinh tương đương vs

    $\cos^6x + \sin^6x – \sin x \cos x = 0$

    $<-> (\cos^2x + \sin^2x) (\cos^4x – \sin^2x \cos^2x + \sin^4x) – \sin x \cos x = 0$

    $<-> \cos^4x + \sin^4x – \sin^2x \cos^2x – \sin x \cos x = 0$

    $<-> (\cos^2x + \sin^2x) – 3\sin^2x \cos^2x – \sin x \cos x = 0$

    $<-> 1 – 3\sin^2x \cos^2x – \sin x \cos x = 0$

    $<-> 1 – \dfrac{3}{4} \sin^2(2x) – \dfrac{1}{2} \sin(2x) = 0$

    Khi đó ta có

    $\sin(2x) = \dfrac{-1 + \sqrt{13}}{3}, \sin(2x) = \dfrac{-1 – \sqrt{13}}{3}$ (loại do $\dfrac{-1 – \sqrt{13}}{3} < -1$)

    Vậy

    $2x = \arcsin(\dfrac{-1 + \sqrt{13}}{3}) + 2k\pi, 2x = \pi – \arcsin(\dfrac{-1 + \sqrt{13}}{3}) + 2k\pi$

    Do đó

    $x = \dfrac{\arcsin(\dfrac{-1 + \sqrt{13}}{3})}{2} + k\pi, x = \dfrac{\pi}{2} – \dfrac{\arcsin(\dfrac{-1 + \sqrt{13}}{3})}{2} + k\pi$.

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