\(2\left(1-sin^23x\right)+cos3x+1=0\)

\(\Leftrightarrow2cos^23x+cos3x+1=0\)

\(\Leftrightarrow2\left(cos3x+\dfrac{1}{4}\right)^2+\dfrac{7}{8}=0\)

Pt đã cho vô nghiệm

Ta có: \(u_n>0\)

Mặt khác: 

\(u_n=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}< \dfrac{1}{n}+\dfrac{1}{n}+...+\dfrac{1}{n}\\ \Leftrightarrow u_n< n.\dfrac{1}{n}=1\)

\(0< u_n< 1\) nên \(u_n\) bị chặn

Xét tính đơn điệu dãy \(u_n\)

\(u_{n+1}=\dfrac{1}{\left(n+1\right)+1}+\dfrac{1}{\left(n+1\right)+2}+...+\dfrac{1}{\left(n+1\right)+\left(n+1-1\right)}+\dfrac{1}{\left(n+1\right)+\left(n+1\right)}\\ =\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}\)

\(u_{n+1}-u_n=\dfrac{1}{2n+1}+\dfrac{1}{2n+2}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}+\dfrac{1}{2\left(n+1\right)}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}-\dfrac{1}{2\left(n+1\right)}\\ =\dfrac{2\left(n+1\right)-\left(2n+1\right)}{2\left(2n+1\right)\left(n+1\right)}=\dfrac{1}{2\left(2n+1\right)\left(n+1\right)}>0\)

\(\Rightarrow u_{n+1}>u_n\)

Dãy \(u_n\) là dãy tăng.

Vậy \(u_n\) bị chặn và tăng nghiêm ngặt nên \(u_n\) hội tụ. đpcm

\(sin3x=sinx\)

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

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

\(\sin3x-\sin x=0\)

\(\Leftrightarrow\sin3x=\sin x\)

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

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

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

\(\Leftrightarrow sin^2x+2sinx.cosx-3cos^2x=1\)

Nhận thấy \(cosx=0\) không phải nghiệm

Với \(cosx\ne0\) chia 2 vế cho \(cos^2x\)

\(\Rightarrow tan^2x+2tanx-3=\dfrac{1}{cos^2x}\)

\(\Leftrightarrow tan^2x+2tanx-3=1+tan^2x\)

\(\Leftrightarrow tanx=2\)

\(\Rightarrow x=arctan\left(2\right)+k\pi\)