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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a/

\(\Leftrightarrow4cos^2\left(6x-2\right)+8\left(1+cos\left(6x-2\right)\right)-13=0\)

Đặt \(cos\left(6x-2\right)=a\Rightarrow\left|a\right|\le1\)

Pt trở thành:

\(4a^2+8\left(1+a\right)-13=0\)

\(\Leftrightarrow4a^2+8a-5=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{2}\\a=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos\left(6x-2\right)=\frac{1}{2}\)

\(\Rightarrow6x-2=\pm\frac{\pi}{3}+k2\pi\)

\(\Rightarrow x=\frac{1}{3}\pm\frac{\pi}{18}+\frac{k\pi}{3}\)

b/

\(\Leftrightarrow2cos^2\left(x+75^0\right)-1+3sin\left(15^0-x\right)-1=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(90^0-15^0+x\right)-2=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(x+75^0\right)-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(x+75^0\right)=\frac{1}{2}\\cos\left(x+75^0\right)=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+75^0=60^0+k360^0\\x+75^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+k360^0\\x=-135^0+k360^0\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x\right)+\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)=1\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow cos2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow1-2sin^2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)\left(1-2sin\left(x-\frac{\pi}{6}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{6}\right)=0\\sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=k\pi\\x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

a/

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

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

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=sin2x\)

\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{\pi}{4}+k2\pi\\2x=\frac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)

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

b/

\(\Leftrightarrow\frac{1-cos2x}{2}+sin2x=\frac{3\left(1+cos2x\right)}{2}\)

\(\Leftrightarrow sin2x-2cos2x=1\)

\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\frac{1}{\sqrt{5}}\)

Đặt \(\frac{1}{\sqrt{5}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Leftrightarrow sin2x.cosa-cos2a.sina=cosa\)

\(\Leftrightarrow sin\left(2x-a\right)=cosa=sin\left(\frac{\pi}{2}-a\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=a-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

c/

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\1-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

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

d/

\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)

\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)

\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)

\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)

a/

\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x-4cosx=0\)

\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx-\sqrt{3}cosx=2\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=1\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=1\)

\(\Leftrightarrow x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{5\pi}{6}+k2\pi\)

b/

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

\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{3}cosx\left(1\right)\\sinx+\sqrt{3}cosx=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow tanx=\sqrt{3}\)

\(\Rightarrow x=\frac{\pi}{3}+k\pi\)

\(\left(2\right)\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin6x\left(cos3x-1-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\Rightarrow x=\frac{k\pi}{6}\\cos3x-sin3x=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin3x-cos3x=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\3x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

\(sin2x=\sqrt{3}cos2x\)

Nhận thấy cos2x=0 ko phải nghiệm, pt tương đương:

\(\frac{sin2x}{cos2x}=\sqrt{3}\Leftrightarrow tan2x=\sqrt{3}\)

\(\Rightarrow2x=\frac{\pi}{3}+k\pi\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

b/

\(cos\left(90^0-x\right)=-sin2x=cos\left(2x+90^0\right)\)

\(\Rightarrow\left[{}\begin{matrix}90^0-x=2x+90^0+k360^0\\90^0-x=-2x-90^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k120^0\\x=-180^0+k360^0\end{matrix}\right.\)

c/ Giống câu a

\(\Leftrightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)

1.

\(\Leftrightarrow sin5x+\sqrt{3}cos5x=-2sin15x\)

\(\Leftrightarrow\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x=-sin15x\)

\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(-15x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=-15x+k2\pi\\5x+\frac{\pi}{3}=\pi+15x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{60}+\frac{k\pi}{10}\\x=-\frac{\pi}{15}+\frac{k\pi}{5}\end{matrix}\right.\)

2.

\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=2\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)=2\)

Do \(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)\le1\\sin\left(x+\frac{\pi}{6}\right)\le1\end{matrix}\right.\) với mọi x

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)\le2\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)=1\\sin\left(x+\frac{\pi}{6}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\\x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

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

\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)

3.

\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)