\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)

\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(PT\Leftrightarrow\left\{{}\begin{matrix}sin3x+cos3x>=0\\2\cdot\left(sin3x+cos3x\right)^2=1+2\cdot sin6x+2\cdot sin2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}sin3x+cos3x>=0\\2+2\cdot sin6x=1+2\cdot sin6x+2\cdot sin2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}sin3x+cos3x>=0\left(1\right)\\sin2x=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)

(2): sin2x=1/2

=>2x=pi/6+k2pi hoặc 2x=5/6pi+k2pi

=>x=pi/12+kpi hoặc x=5/12pi+kpi

Khi x=pi/12+kpi thì:

\(sin3x+cos3x=sin\left(\dfrac{pi}{4}+3\cdot kpi\right)+cos\left(\dfrac{pi}{4}+3\cdot kpi\right)\)

Để sin 3x+cos3x>=0 thì k=2n

Khi x=5/12pi+kpi thì \(sin3x+cos3x=sin\left(\dfrac{5}{4}pi+3\cdot kpi\right)+cos\left(\dfrac{5}{4}pi+3\cdot k\cdot pi\right)\)

Để sin 3x+cos3x>=0 thì \(k=2n+1\)

=>Phương trình ban đầu sẽ có các nghiệm là: \(x=\dfrac{pi}{12}+2npi;x=\dfrac{17}{12}pi+2npi\)

a.\(y'=x\left(\sqrt{x^2-2x}\right)'+\sqrt{x^2-2x}=\dfrac{x}{2\sqrt{x^2-2x}}2\left(x-1\right)+\sqrt{x^2-2x}=\dfrac{x\left(x-1\right)}{\sqrt{x^2-2x}}+\sqrt{x^2-2x}\)

\(=\dfrac{x^2-x+x^2-2x}{2\sqrt{x^2-2x}}=\dfrac{2x^2-3x}{2\sqrt{x^2-2x}}\)

b. \(y=3sin2x+cos3x\Rightarrow y'=6cos2x-3sin3x\)

\(\Rightarrow\sqrt{2}.sin\left(3x-\dfrac{\pi}{4}\right)-\sqrt{2}.sin\left(5x-\dfrac{\pi}{3}\right)=0\Leftrightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(5x-\dfrac{\pi}{3}\right)\)

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

\(\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{19\pi}{96}+\dfrac{k\pi}{4}\end{matrix}\right.\); k\(\in Z\)

Nguyễn Việt Lâm, Nguyễn Lê Phước Thịnh giúp vs!

ĐKXĐ: \(x\ge2\).

Với \(x\ge2\) ta có \(VP\le2;VT\ge2\).

Do đó nghiệm của pt là \(x=2\).

\(\begin{array}{l}a)\;sin2x + cos3x = 0\\ \Leftrightarrow cos\left( {\frac{\pi }{2} - 2x} \right) + cos3x = 0\\ \Leftrightarrow cos\left( {\frac{\pi }{2} - 2x} \right) =  - cos3x\\ \Leftrightarrow cos\left( {\frac{\pi }{2} - 2x} \right) = cos\left( {\pi  - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{\pi }{2} - 2x = \pi  - 3x + k2\pi \\\frac{\pi }{2} - 2x =  - \pi  + 3x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = \frac{{3\pi }}{{10}} + k\frac{{2\pi }}{5}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}b)\;sinx.cosx = \frac{{\sqrt 2 }}{4}\\ \Leftrightarrow \frac{1}{2}\;sin2x = \frac{{\sqrt 2 }}{4}\\ \Leftrightarrow sin2x = \frac{{\sqrt 2 }}{2} = sin\left( {\frac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{4} + k2\pi \\2x = \pi  - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + k\pi \\x = \frac{{3\pi }}{8} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}c)\;sinx + sin2x = 0\\ \Leftrightarrow sinx =  - sin2x\\ \Leftrightarrow sinx = sin( - 2x)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 2x + k2\pi \\x = \pi  + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\frac{{2\pi }}{3}\\x =  - \pi  + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)