đọc lại lý thuyết rồi làm 

1.

Các hàm \(sinx;sin\frac{x}{2};sin\frac{x}{3};...;sin\frac{x}{10}\) có chu kì lần lượt là \(2\pi;4\pi;6\pi;...;20\pi\)

\(\Rightarrow\) Chu kì của hàm đã cho là \(BCNN\left(2\pi;4\pi;...;20\pi\right)=15120\pi\)

2.

a.

\(y=cos^22x+3cos2x+3\)

\(y=\left(cos2x+1\right)\left(cos2x+2\right)+1\ge1\Rightarrow y_{min}=1\) khi \(cos2x=-1\)

\(y=\left(cos2x-1\right)\left(cos2x+4\right)+7\le7\Rightarrow y_{max}=7\) khi \(cos2x=1\)

b.

Đặt \(a=4sinx-3cosx\Rightarrow a^2\le\left(4^2+\left(-3\right)^2\right)\left(sin^2x+cos^2x\right)=25\)

\(\Rightarrow-5\le a\le5\)

\(y=a^2-4a+1\) với \(a\in\left[-5;5\right]\)

\(y=\left(a-2\right)^2-3\ge-3\Rightarrow y_{min}=-3\) khi \(a=2\)

\(y=\left(a-9\right)\left(a+5\right)+46\le46\Rightarrow y_{max}=46\) khi \(a=-5\)

\(y=\left(1-cos^2x\right)^2+cos^2x-5\)

\(y=cos^4x-cos^2x-4\)

\(y=\left(cos^2x-\frac{1}{2}\right)^2-\frac{17}{4}\ge-\frac{17}{4}\)

\(y_{min}=-\frac{17}{4}\) khi \(cos^2x=\frac{1}{2}\)

\(y=cos^2x\left(cos^2x-1\right)-4=-cos^2x.sin^2x-4=-\frac{1}{4}sin^22x-4\)

Do \(-\frac{1}{4}sin^22x\le0\Rightarrow y\le-4\)

\(y_{max}=-4\) khi \(sin2x=0\)

c/

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)

\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)

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

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

\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+sinx-1\right)=0\)

\(\Leftrightarrow2sin^2x+sinx-1=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

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

b/

Nhận thấy \(sinx=0\) không phải nghiệm, pt tương đương:

\(sinx.cosx.cos2x.cos4x.cos8x=\frac{1}{16}sinx\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=sinx\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=sinx\)

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

\(\Rightarrow\left[{}\begin{matrix}16x=x+k2\pi\\16x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{15}\\x=\frac{\pi}{17}+\frac{k2\pi}{17}\end{matrix}\right.\)