ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{\sqrt{3}}{cos^2x}+2+\dfrac{2}{sinx.cosx}-2\sqrt{3}=2\left(\dfrac{1}{tanx}+1\right)\)

\(\Leftrightarrow\sqrt{3}\left(1+tan^2x\right)+\dfrac{\dfrac{2}{cos^2x}}{\dfrac{sinx.cosx}{cos^2x}}+2-2\sqrt{3}=2\left(\dfrac{1}{tanx}+1\right)\)

\(\Leftrightarrow\sqrt{3}tan^2x+\dfrac{2\left(1+tan^2x\right)}{tanx}+2-\sqrt{3}=\dfrac{2}{tanx}+2\)

\(\Leftrightarrow\sqrt{3}tan^3x+2\left(1+tan^2x\right)-\sqrt{3}tanx=2\)

\(\Leftrightarrow\sqrt{3}tan^3x+2tan^2x-\sqrt{3}tanx=0\)

\(\Leftrightarrow...\)

2 cos 2 x   -   3 sin 2 x   +   sin 2 x   =   1

- cosx = 0 thỏa mãn phương trình ⇒ phương trình có nghiệm x = π/2+kπ,k ∈ Z.

- Với cosx ≠ 0, chia hai vế cho cos 2   x , tìm được tanx = 1/6.

Vậy phương trình có các nghiệm x = π/2+kπ,k ∈ Z và x = arctan1/6 + kπ,k ∈ Z.

1   +   sin x   -   cos x   -   sin 2 x   +   2 cos 2 x   =   0   ( 1 )     T a   c ó :     1   -   sin 2 x   =   sin x   -   cos x 2     ⇔   2 cos 2 x   =   2 ( cos 2 x   -   sin 2 x )   =   - 2 ( sin x   -   cos x ) ( sin x   +   cos x )     V ậ y   ( 1 )   ⇔   ( sin x   -   cos x ) ( 1   +   sin x   -   cos x   -   2 sin x   -   2 cos x )   =   0     ⇔   ( sin x   -   cos x ) ( 1   -   sin x   -   3 cos x )   =   0

\(\left(2cos2x-1\right)\left(sin2x+cos2x\right)=1\)

\(\Leftrightarrow2sin2x.cos2x+2cos^22x-sin2x-cos2x-1=0\)

\(\Leftrightarrow sin4x+cos4x-sin2x-cos2x=0\)

\(\Leftrightarrow2cos3x.sinx-2sin3x.sinx=0\)

\(\Leftrightarrow2sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow2\sqrt{2}sinx.cos\left(3x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos\left(3x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\end{matrix}\right.\)

a) <=> 4sinxcosx -(2cos²x-1)=7sinx+2cosx-4

<=> 2cos²x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos²x-2cosx+2=0 

pt trên vn

b) <=> 2sinxcosx-1+2sin²x+3sinx-cosx-1=0

<=> cos(2sinx-1)+2sin²x+3sinx-2=0

<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0

<=> (2sinx-1)(cosx+sinx+2)=0

<=> sinx=1/2 hoặc cosx+sinx=-2(vn)

<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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