\(cos^4x-sin^4x=sin3x+cos4x\)

\(\Leftrightarrow\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=sin3x+cos4x\)

\(\Leftrightarrow cos2x=sin3x+cos4x\)

\(\Leftrightarrow cos4x-cos2x+sin3x=0\)

\(\Leftrightarrow-2sin3x.sinx+sin3x=0\)

\(\Leftrightarrow sin3x\left(1-2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow x=\left\{0;\dfrac{\pi}{3};\dfrac{2\pi}{3};\pi;\dfrac{\pi}{6};\dfrac{5\pi}{6}\right\}\)

\(\Rightarrow\sum x=3\pi\)

\(sinx-\sqrt{3}cosx=1\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{-\dfrac{5\pi}{6};\dfrac{\pi}{2}\right\}\)

A

A

\(sinx.cos3x-sinx+2cos3x-2=0\)

\(\Leftrightarrow sinx\left(cos3x-1\right)+2\left(cos3x-1\right)=0\)

\(\Leftrightarrow\left(sinx+2\right)\left(cos3x-1\right)=0\)

\(\Leftrightarrow cos3x=1\)

\(\Leftrightarrow3x=k2\pi\)

\(\Rightarrow x=\dfrac{k2\pi}{3}\)

Do \(x\in\left(0;2\pi\right)\Rightarrow x=\left\{\dfrac{2\pi}{3};\dfrac{4\pi}{3}\right\}\)

\(\Rightarrow\dfrac{2\pi}{3}+\dfrac{4\pi}{3}=2\pi\)

cos 6x+cos4x=sin7x-sin3x

=>2*cos5x*cosx=2*cos5x*sin2x

=>cos5x(cosx-sin2x)=0

=>cos5x=0 hoặc sin2x=sin(pi/2-x)

=>5x=pi/2+kpi hoặc 2x=pi/2-x+k2pi hoặc 2x=pi/2+x+k2pi

=>x=pi/10+kpi/5; x=pi/6+k2pi/3; x=pi/2+k2pi

Đặt \(t=tan\dfrac{x}{2}\Rightarrow\left\{{}\begin{matrix}t\in\left[0;1\right]\\sinx=\dfrac{2t}{1+t^2}\\cosx=\dfrac{1-t^2}{1+t^2}\end{matrix}\right.\)

Pt trở thành: \(\dfrac{m.2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}=1\)

\(\Leftrightarrow2mt+1-t^2=1+t^2\)

\(\Leftrightarrow2mt-2t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=m\end{matrix}\right.\)

\(\Rightarrow\) Để pt có 2 nghiệm thuộc đoạn đã cho thì \(0< m\le1\)

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

\(\Leftrightarrow-2sin^2x+3sinx+5=0\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{5}{2}\left(l\right)\end{matrix}\right.\)

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

\(0\le-\frac{\pi}{2}+k2\pi\le200\pi\Rightarrow1\le k\le100\) (có 100 nghiệm)

Tổng các nghiệm:

\(\sum x=-\frac{\pi}{2}.100+\sum\limits^{100}_{k=1}2k\pi=10050\pi\)

2.

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

\(\Leftrightarrow2\left|cosx\right|^2+3\left|cosx\right|-2=0\Rightarrow\left[{}\begin{matrix}\left|cosx\right|=\frac{1}{2}\\\left|cosx\right|=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Pt có 2 nghiệm trên đoạn đã cho \(x=\pm\frac{\pi}{3}\)

ĐKXĐ: \(cosx\ne-\dfrac{\sqrt{3}}{2}\) \(\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{5\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(pt\Rightarrow3-\left(1-2sin^2x\right)+2sinx.cosx-5sinx-cosx=0\)

\(\Leftrightarrow2sin^2x-5sinx+2+cosx\left(2sinx-1\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx+cosx=2\left(vn\right)\end{matrix}\right.\) 

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

Loại nghiệm

\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)

\(0\le\dfrac{\pi}{6}+k2\pi\le2022\pi\Rightarrow0\le k\le1010\)

\(\Rightarrow\sum x=1011.\dfrac{\pi}{6}+2\pi\left(0+1+2+...+1010\right)=\dfrac{1011\pi}{6}+2\pi.\dfrac{1010.1011}{2}=...\)