`cos 2x+\sqrt{3}sin 2x+\sqrt{3}sin x-cos x=4`

`<=>1/2 cos 2x+\sqrt{3}/2 sin 2x+\sqrt{3}/2 sin x-1/2 cos x=2`

`<=>sin(\pi/6 +2x)+sin(x-\pi/6)=2`

Vì `-1 <= sin (\pi/6 +2x) <= 1`

     `-1 <= sin (x-\pi/6) <= 1`

 Dấu "`=`" xảy ra `<=>{(sin(\pi/6+2x)=1),(sin(x-\pi/6)=1):}`

        `<=>{(\pi/6+2x=\pi/2+k2\pi),(x-\pi/6=\pi/2+k2\pi):}`

        `<=>{(x=\pi/6+k\pi),(x=[2\pi]/3+k2\pi):}`    `(k in ZZ)`

Pt <=> 2sin\(\dfrac{3x}{2}\).cos\(\dfrac{x}{2}\) = 2cos\(\dfrac{3x}{2}\).cos\(\dfrac{x}{2}\)

⇔ cos\(\dfrac{x}{2}\) . \(\left(sin\dfrac{3x}{2}-cos\dfrac{3x}{2}\right)\) = 0

⇔ \(\sqrt{2}sin\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right).cos\dfrac{x}{2}=0\)

⇔ 

\(cos2x+cosx+1=sin2x+sinx\)

\(\Leftrightarrow cos^2x-sin^2x+cosx+cos^2x+sin^2x=2sinx.cosx+sinx\)

\(\Leftrightarrow2cos^2x+cosx=2sinx.cosx+sinx\)

\(\Leftrightarrow cosx\left(2cosx+1\right)=sinx\left(2cosx+1\right)\)

\(\Leftrightarrow\left(2cosx+1\right)\left(sinx-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2cosx+1=0\\sinx=cosx\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\tanx=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{4}+k\pi\\\end{matrix}\right.\)

Chọn D

Ta sẽ biến đổi phương trình thành dạng tích

Chú ý: có thể dùng 4 đáp án thay vào phương trình để kiểm tra đâu là nghiệm

Chọn A

ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)

\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)

\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)

\(\Leftrightarrow2cos^22x-2cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\2x=k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)

Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)