\(4sin^2x.cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow2\left(1-cos2x\right).cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow2cosx-2cos2x.cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow2cosx-cos3x-cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow\sqrt{3}sin3x+cos3x=2cos2x\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin3x+\dfrac{1}{2}cos3x=cos2x\)

\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos2x\)

\(\Leftrightarrow3x-\dfrac{\pi}{3}=\pm2x+k2\pi\)

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

c/

\(\Leftrightarrow\sqrt{3}sin3x-cos3x=sin2x-\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

e/

\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)

\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)

\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)

Có b nào gipus mk với cần gấp gấp :)

a: \(\frac{1}{\sin x}+\frac{1}{cosx}=4\cdot\sin\left(x+\frac{\pi}{4}\right)\)

=>\(\frac{\sin x+cosx}{\sin x\cdot cosx}=4\cdot\frac{\sqrt2}{2}\cdot\left(\sin x+cosx\right)\)

=>\(\left(\sin x+cosx\right)\left(\frac{1}{\sin x\cdot cosx}-2\sqrt2\right)=0\)

TH1: \(\frac{1}{\sin x\cdot cosx}-2\sqrt2=0\)

=>\(\frac{1}{\sin x\cdot cosx}=2\sqrt2\)

=>\(sinx\cdot cosx=\frac{1}{2\sqrt2}\)

=>\(2\cdot\sin x\cdot cosx=\frac{1}{\sqrt2}\)

=>\(\sin2x=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{4}+k2\pi\\ 2x=-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{8}+k\pi\\ x=-\frac{\pi}{8}+k\pi\end{array}\right.\)

TH2: sin x+cosx=0

=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(x+\frac{\pi}{4}=k\pi\)

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

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

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

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

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

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

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

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

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

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

a: ĐKXĐ: \(\sqrt3\cdot\sin x+cosx<>0\)

=>\(\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx<>0\)

=>\(\sin\left(x+\frac{\pi}{6}\right)<>0\)

=>\(x+\frac{\pi}{6}<>k\pi\)

=>\(x<>-\frac{\pi}{6}+k\pi\)

\(\frac{2\cdot cos2x+1}{\sqrt3\cdot\sin x+cosx}=2\cdot cosx-1\)

=>\(\frac{2\cdot\left(2\cdot cos^2x-1\right)+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}=2\cdot cosx-1\)

=>\(\frac{4\cdot cos^2x-1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-\left(2\cdot cosx-1\right)=0\)

=>\(\left(2\cdot cosx-1\right)\left\lbrack\frac{2\cdot cosx+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-1\right\rbrack=0\)

TH1: \(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}-1=0\)

=>\(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}=1\)

=>\(\sqrt3\cdot\sin x+cosx=2\cdot cosx+1\)

=>\(\sqrt3\cdot\sin x-cosx=1\)

=>\(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx=\frac12\)

=>\(\sin\left(x-\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ x-\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=\pi+k2\pi\end{array}\right.\)

TH2: \(2\cdot cosx-1=0\)

=>\(cosx=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=-\frac{\pi}{3}+k2\pi\end{array}\right.\)