Căn 2(sinx +cos x)=1/2cos2x

Căn 2(sinx +cos x)=1/2cos2x

0 thoughts on “Căn 2(sinx +cos x)=1/2cos2x”

  1. Đáp án: $ x=\dfrac34\pi+k\pi$

    Giải thích các bước giải:

    Ta có:

    $2(\sin x+\cos x)=\dfrac12\cos2x$

    $\to 2(\sin x+\cos x)=\dfrac12(\cos^2x-\sin^2x)$

    $\to 2(\sin x+\cos x)=\dfrac12(\cos x-\sin x)(\cos x+\sin x)$

    $\to 4(\sin x+\cos x)=(\cos x-\sin x)(\cos x+\sin x)$

    $\to 4(\sin x+\cos x)-(\cos x-\sin x)(\cos x+\sin x)=0$

    $\to (\sin x+\cos x)+(4-\cos x+\sin x)=0$

    Vì $-1\le \sin x,\cos x\le 1$

    $\to 4-\cos x+\sin x\ge 4-1-1=2>0$

    $\to \sin x+\cos x=0$

    $\to \sin x=-\cos x$

    Nếu $\cos x=0\to \sin x=0$ vô lý vì $\sin^2x+\cos^2x=1$

    $\to \cos x\ne 0$

    $\to \dfrac{\sin x}{\cos x}=-1$

    $\to \tan x=-1$

    $\to x=\dfrac34\pi+k\pi$

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  2. $\sqrt2(\sin x+\cos x)=\dfrac{1}{2}(\cos2x)$

    $\Leftrightarrow \sqrt2(\sin x+\cos x)-\dfrac{1}{2}(\cos^2x-\sin^2x)=0$

    $\Leftrightarrow \sqrt2(\sin x+\cos x)+\dfrac{1}{2}(\sin x+\cos x)(\sin x-\cos x)=0$

    $\Leftrightarrow (\sin x+\cos x)(\sqrt2+\dfrac{1}{2}(\sin x-\cos x))=0$

    $+) \sin x+\cos x=0\Leftrightarrow \sqrt2\sin(x+\dfrac{\pi}{4})=0\Leftrightarrow x=\dfrac{-\pi}{4}+k\pi$

    $+) \sqrt2-\dfrac{1}{2}(\sin x-\cos x)=0\Leftrightarrow \sin x-\cos x=2\sqrt2\Leftrightarrow \sqrt2\sin(x-\dfrac{\pi}{4})=2\sqrt2\Leftrightarrow \sin(x-\dfrac{\pi}{4})=2>1$ (loại)

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