Giup e 2 ý luôn ạ, e cảm ơn mnn Giải pt a.cos2x/1-tanx=0 b.sin2x/(1-cos2x)=0

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Giup e 2 ý luôn ạ, e cảm ơn mnn
Giải pt
a.cos2x/1-tanx=0
b.sin2x/(1-cos2x)=0

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Faith 2 tuần 2021-08-28T07:53:57+00:00 2 Answers 0 views 0

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    2021-08-28T07:55:25+00:00

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    2021-08-28T07:55:25+00:00

    Đáp án: a.$x=\dfrac34\pi+k\pi,k\in Z$

                 b.$x=\dfrac12\pi+k\pi$

    Giải thích các bước giải:

    a.ĐKXĐ: $x\ne \frac{\pi }{4}+k\pi, x\ne \dfrac{\pi}{2}+k\pi$

    Ta có:

    $\dfrac{\cos2x}{1-\tan x}=0$

    $\to \cos2x=0$

    $\to 2x=\dfrac{\pi}{2}+k\pi$

    $\to x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$

    Mà $x\ne \frac{\pi }{4}+k\pi$

    $\to x=\dfrac{3\pi}{4}+k\pi$

    b.ĐKXĐ: $x\ne k\pi$

    Ta có:

    $\dfrac{\sin2x}{1-\cos2x}=0$

    $\to \sin2x=0$

    $\to \cos^22x=1-\sin^22x=1$

    $\to \cos2x=-1$ vì $\cos2x\ne 0$

    $\to 2x=\pi+k2\pi$

    $\to x=\dfrac12\pi+k\pi$

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