1. \(D=R\)

2. 

Hàm số xác định khi:

\(\left\{{}\begin{matrix}cos5x\ne0\\sin4x-cos3x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x\ne\dfrac{\pi}{2}+k\pi\\cos\left(\dfrac{\pi}{2}-4x\right)-cos3x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{10}+\dfrac{k\pi}{5}\\sin\left(\dfrac{\pi}{4}-\dfrac{7x}{2}\right).sin\left(\dfrac{\pi}{4}-x\right)\ne0\end{matrix}\right.\)

...

3.

Hàm số xác định khi:

\(1+cosx\ne0\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)

87.

\(\left(sinx+\sqrt{3}cosx\right)sin3x=2\)

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

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right).sin3x=1\)

\(\Leftrightarrow\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{3}\right)-\dfrac{1}{2}cos\left(4x+\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(4x+\dfrac{\pi}{3}\right)=2\)

Ta có:

\(cos\left(2x-\dfrac{\pi}{3}\right)\le1;cos\left(4x+\dfrac{\pi}{3}\right)\ge-1\)

\(\Rightarrow cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(4x+\dfrac{\pi}{3}\right)\le2\)

Đẳng thức xảy ra khi: 

\(\left\{{}\begin{matrix}cos\left(2x-\dfrac{\pi}{3}\right)=1\\cos\left(4x+\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{\pi}{3}=k2\pi\\4x+\dfrac{\pi}{3}=\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Câu 88:

\(\sqrt{5+sin^2x}=sinx+2cosx\)

\(\Leftrightarrow\sqrt{1+\dfrac{sin^2x}{5}}=\dfrac{1}{\sqrt{5}}sinx+\dfrac{2}{\sqrt{5}}cosx\)

\(\Leftrightarrow\sqrt{1+\dfrac{sin^2x}{5}}=sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)\)

Có \(\sqrt{1+\dfrac{sin^2x}{5}}\ge\sqrt{1+0}=1\)

\(sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)\le1\)

Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=0\\sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x+arc.cos\dfrac{1}{\sqrt{5}}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x=k\pi-arc.cos\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy \(S=\varnothing\)

d, \(cosx-cos2x=sin3x\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}=2sin\dfrac{3x}{2}.cos\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}-cos\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\sin\dfrac{x}{2}=cos\dfrac{3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\cos\left(\dfrac{\pi}{2}-\dfrac{x}{2}\right)=cos\dfrac{3x}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{4}-k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

Đăng tầm 2, 3 câu 1 lần thôi.

g, \(1+2sinx=2cosx\)

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

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x-\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

h, \(4cosx-3sinx=3\)

\(\Leftrightarrow5\left(\dfrac{4}{5}cosx-\dfrac{3}{5}sinx\right)=3\)

\(\Leftrightarrow cos\left(x+arccos\dfrac{4}{5}\right)=\dfrac{3}{5}\)

\(\Leftrightarrow x+arccos\dfrac{4}{5}=\pm arccos\dfrac{3}{5}+k2\pi\)

\(\Leftrightarrow x=-arccos\dfrac{4}{5}\pm arccos\dfrac{3}{5}+k2\pi\)

a, \(sin5x-cos5x+1=0\)

\(\Leftrightarrow\sqrt{2}sin\left(5x-\dfrac{\pi}{4}\right)+1=0\)

\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\5x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{5}\\x=\dfrac{3\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

b, \(sin3x+cos3x=1\)

\(\Leftrightarrow\sqrt{2}sin\left(3x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

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

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

\(y=sin^2\left(x-\dfrac{\pi}{5}\right)+sin\left(x-\dfrac{\pi}{5}\right)\)

Đặt \(t=sin\left(x-\dfrac{\pi}{5}\right)\left(t\in\left[-1;1\right]\right)\)

Khi đó: \(y=f\left(t\right)=t^2+t\)

\(y_{min}=min\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{1}{2}\right)\right\}=-\dfrac{1}{4}\)

\(y_{max}=max\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{1}{2}\right)\right\}=2\)

Có thể lập \(C_4^2=6\) tập con có chứa 2 phần tử.

ta có :

\(\hept{\begin{cases}cosX\ne0\\\sqrt{3}sinX+cosX\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\frac{\pi}{2}+k2\pi\\tanX\ne-\frac{1}{\sqrt{3}}\Leftrightarrow x\ne-\frac{\pi}{6}+k\pi\end{cases}}\)

\(Y=\frac{tanX}{\sqrt{3}sinX+cosX}\)\(\left(ĐKXĐ:\hept{\begin{cases}cosX\ne0\\\sqrt{3}sinX+cosX\ne0\end{cases}}\right)\)

\(\Leftrightarrow\hept{\begin{cases}X\ne\frac{\pi}{2}+k\pi\\\frac{\sqrt{3}}{2}sinX+\frac{1}{2}cosX\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}X\ne\frac{\pi}{2}+k\pi\\sin\left(X+\frac{\pi}{6}\right)\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}X\ne\frac{\pi}{2}+k\pi\\X+\frac{\pi}{6}\ne k\pi\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}X\ne\frac{\pi}{2}+k\pi\\X\ne-\frac{\pi}{6}+k\pi\end{cases}}\) \(\left(k\inℤ\right)\)

\(TXĐ:D\)

\(=ℝ\backslash\left\{\frac{\pi}{2}+k\pi;-\frac{\pi}{6}+k\pi\right\}\)