Môn Toán Lớp 11: Giải Phương trình a, sin2x – (căn 2)cosx = 0 b, cos2x + cosx =sinx

a)sin 2x-\sqrt{2} cos x=0

<=>2sin x cos x-\sqrt{2} cos x=0

<=>\sqrt{2}cos x(\sqrt{2} sin x-1)=0

<=> $\left[\begin{matrix} \sqrt{2} cos x=0\\ \sqrt{2} sin x-1=\end{matrix}\right.$ 

<=> $\left[\begin{matrix} cos x=0\\ sin x=\dfrac{\sqrt{2}}{2}\end{matrix}\right.$

<=> $\left[\begin{matrix} x=\dfrac{\pi}{2}+k\pi\\ x=\dfrac{\pi}{4}+k2\pi\\ x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.$    (k in ZZ)

Vậy S={\pi/2+k\pi;\pi/4+k2\pi;[3\pi]/4+k2\pi|k in ZZ}

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b)cos 2x+cos x=sin x

<=>cos^2 x-sin^2 x+cos x-sin x=0

<=>(cos x-sin x)(cos x+sin x)+(cos x-sin x)=0

<=>(cos x-sin x)(cos x+sin x+1)=0

@TH1:cos x-sin x=0

    <=>1/\sqrt{2}cos x-1/\sqrt{2}sin x=0

    <=>cos(x+\pi/4)=0

    <=>x+\pi/4=\pi/2+k\pi

    <=>x=\pi/4+k\pi   (k in ZZ)

@TH2:cos x+sin x+1=0

    <=>cos x+sin x=-1

    <=>1/\sqrt{2}cos x+1/\sqrt{2}sin x=-1/\sqrt{2}

    <=>cos(x-\pi/4)=-1/\sqrt{2}

    <=> $\left[\begin{matrix} x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\\ x-\dfrac{\pi}{4}=\dfrac{-3\pi}{4}+k2\pi\end{matrix}\right.$

    <=> $\left[\begin{matrix} x=\pi+k2\pi\\ x=\dfrac{-\pi}{2}+k2\pi\end{matrix}\right.$   (k in ZZ)

Vậy S={\pi/4+k\pi;\pi+k2\pi;[-\pi]/2+k2\pi|k in ZZ}