Vì 1 = 5 °  nên ta có

Nhận thấy \(sinx=0\) ko phải nghiệm

\(\Leftrightarrow sinx\left(2cos2x+1\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(2cos2x.sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(sin3x-sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(2cos6x.sin3x+sin3x\right)\left(2cos18x+1\right)=sinx\)

\(\Leftrightarrow\left(2cos18x.sin9x+sin9x\right)=sinx\)

\(\Leftrightarrow sin27x=sinx\)

\(sinx\left(1+2cos2x+2cos4x+2cos6x\right)\)

\(=sinx+2sinx.cos2x+2sinx.cos4x+2sinx.cos6x\)

\(=sinx+sin3x+sin\left(-x\right)+sin5x+sin\left(-3x\right)+sin7x+sin\left(-5x\right)\)

\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)

\(=sin7x\)

a/ \(y=13\left(\frac{12}{13}sin3x+\frac{5}{13}cos3x\right)\)

Đặt \(cosa=\frac{12}{13}\) với \(a\in\left(0;\pi\right)\Rightarrow y=13\left(sin3x.cosa+cos3x.sina\right)=13sin\left(3x+a\right)\)

\(\Rightarrow-13\le y\le13\)

\(y_{min}=-13\) khi \(sin\left(3x+a\right)=-1\)

\(y_{max}=13\) khi \(sin\left(3x+a\right)=1\)

Hoặc bạn cũng có thể dùng BĐT Bunhiacopxki, tùy

b/

\(x\in\left(\pi+k2\pi;\frac{3\pi}{2}+k2\pi\right)\Rightarrow tanx>0\)

\(y=\frac{tanx}{2}+\frac{tanx}{2}+\frac{1}{tan^2x}\ge3\sqrt[3]{\frac{tan^2x}{4tan^2x}}=\frac{3}{\sqrt[3]{4}}\)

\(y_{min}=\frac{3}{\sqrt[3]{4}}\) khi \(tanx=\sqrt[3]{2}\)

\(y_{max}\) ko tồn tại

\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)

\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)

\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)

\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)

\(=sin7x\)

1.

Chọn 3 chữ số còn lại từ 7 chữ số còn lại: \(C_7^3=35\) cách

Hoán vị 5 chữ số: \(5!=120\)

Số số thỏa mãn: \(35.120=4200\) số

2.

a.

\(\sqrt{3}sin5x-2sin5x.cos5x=0\)

\(\Leftrightarrow sin5x\left(\sqrt{3}-2cos5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\cos5x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{5}\\x=\pm\frac{\pi}{30}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=cos6x\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos6x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{24}+\frac{k\pi}{4}\end{matrix}\right.\)