Tìm nguyên hàm sau:
$\displaystyle\int
\left(3x^2 - \frac{4}{x} + \sin3x - \cos4x + e^{2x+1} + 3^{2x-2} + 3\sqrt{x^4} + \frac{1}{\cos^2x} - \frac{1}{\sin^2x}\right) dx$
Tìm các nguyên hàm sau:
a) \(\int (3x^2-2x-4)dx \)
b) \(\int(\sin3x-\cos4x)dx \)
c) \(\int(e^{-3x}-4^x)dx \)
d) \(\int\ln(x)dx \)
e) \(\int(x.e^x)dx \)
f) \(\int(x+1).\sin(x)dx \)
g) \(\int x.\ln(x)dx \)
\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)
\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)
\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)
d. \(I=\int lnxdx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x\end{matrix}\right.\)
\(\Rightarrow u=x.lnx-\int dx=x.lnx-x+C\)
e. Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=x.e^x-\int e^xdx=x.e^x-e^x+C\)
f.
Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)
\(\Rightarrow I=-\left(x+1\right)cosx+\int cosxdx=-\left(x+1\right)cosx+sinx+C\)
g.
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)
Tính các nguyên hàm sau :
a) \(\int x\left(3-x\right)^5dx\)
b) \(\int\left(2^x-3^x\right)^2dx\)
c) \(\int x\sqrt{2-5x}dx\)
d) \(\int\dfrac{\ln\left(\cos x\right)}{\cos^2x}dx\)
e) \(\int\dfrac{x}{\sin^2x}dx\)
\(\int\dfrac{x+1}{\left(x-2\right)\left(x+3\right)}dx\)
h) \(\int\dfrac{1}{1-\sqrt{x}}dx\)
i) \(\int\sin3x\cos2xdx\)
k) \(\int\dfrac{\sin^3x}{\cos^2x}dx\)
l) \(\int\dfrac{\sin x\cos x}{\sqrt{a^2\sin^2x+b^2\cos^2x}}dx\) (\(a^2\ne b^2\))
Tìm các nguyên hàm sau:
a) \(I_1=\int\frac{\left(x^2+3\right)dx}{\sqrt{\left(2x-5\right)^3}}\)
b)\(I_2=\int\frac{dx}{\left(3x-1\right)\ln\left(3x-1\right)}\)
c) \(I_3=\int\frac{\left(x^2+1\right)dx}{\sqrt{x^6-7x^4+x^2}}\)
a) Đặt \(\sqrt{2x-5}=t\) khi đó \(x=\frac{t^2+5}{2}\) , \(dx=tdt\)
Do vậy \(I_1=\int\frac{\frac{1}{4}\left(t^2+5\right)^2+3}{t^3}dt=\frac{1}{4}\int\frac{\left(t^4+10t^2+37\right)t}{t^3}dt\)
\(=\frac{1}{4}\int\left(t^2+10+\frac{37}{t^2}\right)dt=\frac{1}{4}\left(\frac{t^3}{3}+10t-\frac{37}{t}\right)+C\)
Trở về biến x, thu được :
\(I_1=\frac{1}{12}\sqrt{\left(2x-5\right)^3}+\frac{5}{2}\sqrt{2x-5}-\frac{37}{4\sqrt{2x-5}}+C\)
b) \(I_2=\frac{1}{3}\int\frac{d\left(\ln\left(3x-1\right)\right)}{\ln\left(3x-1\right)}=\frac{1}{3}\ln\left|\ln\left(3x-1\right)\right|+C\)
c) \(I_3=\int\frac{1+\frac{1}{x^2}}{\sqrt{x^2-7+\frac{1}{x^2}}}dx=\int\frac{d\left(x-\frac{1}{x}\right)}{\sqrt{\left(x-\frac{1}{2}\right)^2-5}}\)
Đặt \(x-\frac{1}{x}=t\)
\(\Rightarrow\) \(I_3=\int\frac{dt}{\sqrt{t^2-5}}=\ln\left|t+\sqrt{t^2-5}\right|+C\)
\(=\ln\left|x-\frac{1}{x}+\sqrt{x^2-7+\frac{1}{x^2}}\right|+C\)
Tìm nguyên hàm các hàm số lượng giác sau :
a) \(\int\frac{dx}{\cos^2x\sin^2x}\) b) \(\int\left(\tan x+\cot x\right)^2dx\)
c) \(\int\tan^2xdx\) d) \(\int\left(5^{3x}+\frac{1}{\sin^2\left(2x+1\right)}+\frac{1}{\sqrt[5]{4x-1}}\right)dx\)
a) Áp dụng đồng nhất thức \(\cos^2x+\sin^2x=1\)
ta có : \(\int\frac{1}{\cos^2x.\sin^2x}dx=\int\frac{\cos^2x+\sin^2x}{\cos^2x.\sin^2x}dx=\int\frac{dx}{\sin^2x}+\int\frac{dx}{\cos^2x}\)
\(=-\cot x+\tan x+C\)
b) Khai triển biểu thức dưới dấu nguyên hàm ta thu được :
\(\int\left(\tan x+\cot x\right)^2dx=\int\left(\tan^2x+2+\cot^2x\right)dx\)
\(=\int\left[\left(\tan^2x+1\right)+\left(\cot^2x+1\right)\right]dx\)
\(=\int\frac{dx}{\cos^2x}+\int\frac{dx}{\sin^2x}\)
\(=\tan x-\cot x+C\)
c) \(\int\tan^2xdx=\int\left(\frac{1}{\cos^2x}-1\right)dx=\tan x-x+C\)
d) \(\int\left(5^{3x}+\frac{1}{\sin^2\left(2x+1\right)}+\frac{1}{\sqrt[5]{4x-1}}\right)dx=\)
\(=\int5^{3x}dx+\int\frac{dx}{\sin^2\left(2x+1\right)}+\int\frac{dx}{\sqrt[5]{4x-1}}\)
\(=\frac{1}{3}\int5^{3x}d\left(3x\right)+\frac{1}{2}\int\frac{d\left(2x+1\right)}{\sin^2\left(2x+1\right)}+\frac{1}{4}\int\left(4x-1\right)^{-\frac{1}{5}}d\left(4x-1\right)\)
\(=\frac{5^{3x}}{3\ln5}-\frac{1}{2}\cot\left(2x+1\right)+\frac{5}{16}\sqrt[5]{\left(4x-1\right)^4+C}\)
1) \(\int\frac{xdx}{1+\sqrt{x-1}}\)
2) \(\int\frac{sin2xdx}{\cos^3x-\sin^2x-1}\)
3) \(\int\frac{dx}{1+\sqrt{x}+\sqrt{1+x}}\)
4) \(\int\frac{dx}{3x^3+x^2-4x}\)
5) \(\int\frac{dx}{\sqrt{9-x^2}}\)
1) Đặt \(t=1+\sqrt{x-1}\Leftrightarrow x=\left(t-1\right)^2+1\forall t\ge1\Rightarrow dx=d\left(t-1\right)^2=2dt\)
\(\Rightarrow I_1=\int\frac{\left(t-1\right)^2+1}{t}\cdot2dt=2\int\frac{t^2-2t+2}{t}dt=2\int\left(t-2+\frac{2}{t}\right)dt\\ =t^2-4t+4lnt+C\)
Thay x vào ta có...
2) \(I_2=\int\frac{2sinx\cdot cosx}{cos^3x-\left(1-cos^2x\right)-1}dx=\int\frac{-2cosx\cdot d\left(cosx\right)}{cos^3x+cos^2x-2}=\int\frac{-2t\cdot dt}{t^3+t-2}\)
\(I_2=\int\frac{-2t}{\left(t-1\right)\left(t^2+2t+2\right)}dt=-\frac{2}{5}\int\frac{dt}{t-1}+\frac{1}{5}\int\frac{2t+2}{t^2+2t+2}dt-\frac{6}{5}\int\frac{dt}{\left(t+1\right)^2+1}\)
Ta có:
\(\int\frac{2t+2}{t^2+2t+2}dt=\int\frac{d\left(t^2+2t+2\right)}{t^2+2t+2}=ln\left(t^2+2t+2\right)+C\)
\(\int\frac{dt}{\left(t+1\right)^2+1}=\int\frac{\frac{1}{cos^2m}}{tan^2m+1}dm=\int dm=m+C=arctan\left(t+1\right)+C\)
Thay x vào, ta có....
3)
\(\frac{1}{\left(1+\sqrt{x}\right)+\sqrt{x+1}}=\frac{\left(1+\sqrt{x}\right)-\sqrt{x+1}}{\left[\left(1+\sqrt{x}\right)-\sqrt{x+1}\right]\cdot\left[\left(1+\sqrt{x}\right)+\sqrt{x+1}\right]}\\ =\frac{\left(1+\sqrt{x}\right)-\sqrt{x+1}}{2\sqrt{x}}=\frac{1}{2\sqrt{x}}+\frac{1}{2}+\frac{\sqrt{x+1}}{2\sqrt{x}}\)
\(I_3=\int\left(\frac{1}{2\sqrt{x}}+\frac{1}{2}+\frac{\sqrt{x+1}}{2\sqrt{x}}\right)dx=\sqrt{x}+\frac{x}{2}+\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}\)
Xét \(\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}\)
Đặt \(x=tan^2t\Leftrightarrow dx=\frac{2tant}{cos^2t}\cdot dt\)
\(\Rightarrow\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}=\int\sqrt{\frac{tan^2t+1}{tan^2t}}\cdot\frac{tant}{cos^2t}dt\\ =\int\frac{1}{sin^2t}\cdot\frac{sint}{cos^3t}dt=\int\frac{d\left(cost\right)}{cos^3t\left(1-cos^2t\right)}=...\)
1)\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)
2)\(\int\frac{dx}{\left(e^x+1\right)\left(x^2+1\right)}\)
3)\(\int\frac{1+2x\sqrt{1-x^2}+2x^2}{1+x+\sqrt{1+x^2}}\)dx
4)\(\int\frac{sin^6x+c\text{os}^6x}{1+6^x}dx\)
5)\(\int_0^{\frac{\pi}{2}}\frac{\sqrt{c\text{os}x}}{\sqrt{s\text{inx}}+\sqrt{c\text{os}x}}dx\)
6)\(\int\frac{x^4}{2^x+1}dx\)
7)\(\int_0^{\frac{\pi^2}{4}}sin\sqrt{x}dx\)
8)\(\int\sqrt[6]{1-c\text{os}^3x}.s\text{inx}.c\text{os}^5xdx\)
9)\(\int\sqrt{\frac{1}{4x}+\frac{\sqrt{x}+e^x}{\sqrt{x}.e^x}}dx\)
10)\(\int\frac{c\text{os}x+s\text{inx}}{\left(e^xs\text{inx}+1\right)s\text{inx}}dx\)
Giải các phương trình sau:
a) \(\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\)
b) \(\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\)
c) \(\sin 3x - \cos 5x = 0\)
d) \({\cos ^2}x = \frac{1}{4}\)
e) \(\sin x - \sqrt 3 \cos x = 0\)
f) \(\sin x + \cos x = 0\)
a)
\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
b) \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
d)
\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)
e)
\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)
f)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
Tìm các nguyên hàm sau
1.\(\int\frac{9x^2}{\sqrt{1-x^3}}dx\)
2.\(\int\frac{1}{\sqrt{x}\left(1+\sqrt{x}\right)^3}dx\)
3.\(\int\frac{x}{\sqrt{2x+3}}dx\)
4.\(\int\) \(\frac{e^{2x}}{\sqrt{1+e^x}}\) dx
5.\(\int\frac{\sqrt[3]{1+lnx}}{x}dx\)
6.\(\int\) cosxsin3xdx
7.\(\int\) (x2+2x-1)exdx
8.\(\int\) excosxdx
9.\(\int\) xsin(2x+1)dx
10.\(\int\) (1-2x)e3xdx
Không phải tất cả các câu đều dùng nguyên hàm từng phần được đâu nhé, 1 số câu phải dùng đổi biến, đặc biệt những câu liên quan đến căn thức thì đừng dại mà nguyên hàm từng phần (vì càng nguyên hàm từng phần biểu thức nó càng phình to ra chứ không thu gọn lại, vĩnh viễn không ra kết quả đâu)
a/ \(I=\int\frac{9x^2}{\sqrt{1-x^3}}dx\)
Đặt \(u=\sqrt{1-x^3}\Rightarrow u^2=1-x^3\Rightarrow2u.du=-3x^2dx\)
\(\Rightarrow9x^2dx=-6udu\)
\(\Rightarrow I=\int\frac{-6u.du}{u}=-6\int du=-6u+C=-6\sqrt{1-x^3}+C\)
b/ Đặt \(u=1+\sqrt{x}\Rightarrow du=\frac{dx}{2\sqrt{x}}\Rightarrow2du=\frac{dx}{\sqrt{x}}\)
\(\Rightarrow I=\int\frac{2du}{u^3}=2\int u^{-3}du=-u^{-2}+C=-\frac{1}{u^2}+C=-\frac{1}{\left(1+\sqrt{x}\right)^2}+C\)
c/ Đặt \(u=\sqrt{2x+3}\Rightarrow u^2=2x\Rightarrow\left\{{}\begin{matrix}x=\frac{u^2}{2}\\dx=u.du\end{matrix}\right.\)
\(\Rightarrow I=\int\frac{u^2.u.du}{2u}=\frac{1}{2}\int u^2du=\frac{1}{6}u^3+C=\frac{1}{6}\sqrt{\left(2x+3\right)^3}+C\)
d/ Đặt \(u=\sqrt{1+e^x}\Rightarrow u^2-1=e^x\Rightarrow2u.du=e^xdx\)
\(\Rightarrow I=\int\frac{\left(u^2-1\right).2u.du}{u}=2\int\left(u^2-1\right)du=\frac{2}{3}u^3-2u+C\)
\(=\frac{2}{3}\sqrt{\left(1+e^x\right)^2}-2\sqrt{1+e^x}+C\)
e/ Đặt \(u=\sqrt[3]{1+lnx}\Rightarrow u^3=1+lnx\Rightarrow3u^2du=\frac{dx}{x}\)
\(\Rightarrow I=\int u.3u^2du=3\int u^3du=\frac{3}{4}u^4+C=\frac{3}{4}\sqrt[3]{\left(1+lnx\right)^4}+C\)
f/ \(I=\int cosx.sin^3xdx\)
Đặt \(u=sinx\Rightarrow du=cosxdx\)
\(\Rightarrow I=\int u^3du=\frac{1}{4}u^4+C=\frac{1}{4}sin^4x+C\)
Từ phần này trở đi mới bắt đầu xài nguyên hàm từng phần:
g/ \(I=\int\left(x^2+2x-1\right)e^xdx\)
Đặt \(\left\{{}\begin{matrix}u=x^2+2x-1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\left(2x+2\right)dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+2x-1\right)e^x-\int\left(2x+2\right)e^xdx\)
Xét \(J=\int\left(2x+2\right)e^xdx\)
Đặt \(\left\{{}\begin{matrix}u=2x+2\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow J=\left(2x+2\right)e^x-\int2e^xdx=\left(2x+2\right)e^x-2e^x+C=2x.e^x+C\)
\(\Rightarrow I=\left(x^2+2x-1\right)e^x-2x.e^x+C=\left(x^2-1\right)e^x+C\)
giải phương trình
a, \(2\sin\frac{x}{2}\left(\sin\frac{3x}{2}+\cos\frac{3x}{2}\right)=3-4\cos x\)
b, \(\frac{2\cos^2x+\sqrt{3}\sin2x+3}{2\cos^2x.\sin\left(x+\frac{\pi}{3}\right)}=3\left(\tan^2x+1\right)\)
a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)
\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)
\(\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)
\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)
\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)