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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Đk: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m2\pi\\x\ne\dfrac{\pi}{4}+n\pi\end{matrix}\right.\left(m,n\in Z\right)\)

PT \(\Leftrightarrow1=2\sqrt{2}sinx.cosx\left(sinx-cosx\right)+2cos^2x\)

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

\(\Leftrightarrow\sqrt{2}sin2x\left(sinx-cosx\right)+\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)

\(\Leftrightarrow\sqrt{2}sin2x=sinx+cosx\)

\(\Leftrightarrow\sqrt{2}sin2x=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\pi-x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+k\dfrac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

ĐKXĐ: \(sinx\ne0\)

\(2cos^2x-3cosx+1=sinx-2sinx^2cosx+2cos^2x.sinx\)

\(\Leftrightarrow2cos^2x\left(1-sinx\right)+1-sinx-3cosx+2sin^2x.cosx=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(2cos^2x+1\right)-cosx\left(3-2sin^2x\right)=0\)

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

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

\(\Leftrightarrow sinx+cosx=1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

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

a ) \(2cosx-3sinx+2=0\) 

\(\Leftrightarrow2cosx-3sinx=-2\)  

\(\Leftrightarrow\dfrac{2}{\sqrt{13}}cosx-\dfrac{3}{\sqrt{13}}sinx=-\dfrac{2}{\sqrt{13}}\) 

Thấy : \(\left(\dfrac{2}{\sqrt{13}}\right)^2+\left(\dfrac{-3}{\sqrt{13}}\right)^2=1\) nên tồn tại \(\alpha\) t/m : 

\(sin\alpha=\dfrac{2}{\sqrt{13}};cos\alpha=\dfrac{-3}{\sqrt{13}}\) . . Khi đó : \(sin\alpha.cosx+cos\alpha.sinx=\dfrac{-2}{\sqrt{13}}\)

\(\Leftrightarrow sin\left(\alpha+x\right)=\dfrac{-2}{\sqrt{13}}\) ( p/t cơ bản ) 

b ) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\) ( ĐK : \(cosx\ne-1\Leftrightarrow x\ne\left(2k+1\right)\pi\) ; ( k thuộc Z )  ) 

\(\Leftrightarrow2+2sinx=cosx+1\) \(\Leftrightarrow cosx-2sinx=1\) 

Làm giống như a )  

c ) \(cos\left(2x-15^o\right)+sin\left(2x-15^o\right)=-1\)  

Đặt \(t=2x-15^o\) ; ta có : \(cos t + sin t = -1\) 

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

Xong rồi bn làm tiếp ; chú ý đổi ra độ 

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

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

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

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

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

Nghiệm dương nhỏ nhất là \(x=\frac{\pi}{3}\)