giải pt 20/ $\frac{1+sin2x+cos2x}{1+cot^2x}= √2sinx.sin2x$

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giải pt
20/ $\frac{1+sin2x+cos2x}{1+cot^2x}= √2sinx.sin2x$

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Dâu 4 years 2020-10-30T12:39:23+00:00 1 Answers 90 views 0

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  1. Ben
    0
    2020-10-30T12:40:30+00:00
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    ĐK: $\sin x\ne 0\Leftrightarrow x\ne k\pi$

    $\dfrac{1+\sin2x+2\cos^2x-1}{\dfrac{1}{\sin^2x}}$

    $=\sin^2x(2\cos^2x+\sin2x)$

    $\Rightarrow \sin^2x(2\cos^2x+\sin2x)-\sqrt2\sin x.\sin2x=0$

    $\Leftrightarrow \sin^2x(2\cos^2x+\sin2x-2\sqrt2\cos x)=0$

    $\Leftrightarrow 2\sin^2x(\cos^2x+\sin x\cos x+\sqrt2\cos x)=0$

    $\Leftrightarrow 2\sin^2x\cos x(\cos x+\sin x+\sqrt2)=0$

    $+) \sin x=0\Leftrightarrow x=k\pi$ (loại)

    $+) \cos x=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi$ (TM)

    $+) \cos x+\sin x=-\sqrt2\Leftrightarrow \sqrt2\sin(x+\dfrac{\pi}{4})=-\sqrt2\Leftrightarrow \sin(x+\dfrac{\pi}{4})=-1\Leftrightarrow x=\dfrac{-3\pi}{4}+k2\pi$ (TM)

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