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}`