Giải pt: a, 5(sinx-cosx)^2=cos4x+2 b, 2sin^2cos^2x-sin2x+1/2=0

Giải pt:
a, 5(sinx-cosx)^2=cos4x+2
b, 2sin^2cos^2x-sin2x+1/2=0

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  1. a) $5(sinx-cosx)^2=cos4x+2$

    $↔ 5(1-sin2x)=1-2sin^22x+2$

    $↔ 5-5sin2x=1-2sin^22x+2$

    $↔ 2sin^22x-5sin2x+2=0$

    $↔ \left[ \begin{array}{l}sin2x=2\\sin2x=\dfrac{1}{2}\end{array} \right.$

    Vì $sin2x∈[-1;1]$ nên loại $sin2x=2$

    $sin2x=\dfrac{1}{2}$

    $↔ sin2x=sin\dfrac{\pi}{6}$

    $↔ \left[ \begin{array}{l}2x=\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.$

    $↔ \left[ \begin{array}{l}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\end{array} \right.$ $(k∈Z)$

    b) $2sin^2xcos^2x-sin2x+\dfrac{1}{2}=0$

    $↔ 2\Bigg(\dfrac{1}{2}sin2x\Bigg)^2-sin2x+\dfrac{1}{2}=0$

    $↔ \dfrac{1}{2}sin^22x-sin2x+\dfrac{1}{2}=0$

    $↔ sin^22x-2sin2x+1=0$

    $↔ (sin2x-1)^2=0$

    $↔ sin2x-1=0$

    $↔ sin2x=1$

    $↔ 2x=\dfrac{\pi}{2}+k2\pi$

    $↔ x=\dfrac{\pi}{4}+k\pi$ $(k∈Z)$

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