\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)

\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)

\(F=\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)

\(F=\frac{2sin4xcos3x+sin4x}{2cos4xcos3x+cos4x}\)

\(F=\frac{2sin4x\left(cos3x+1\right)}{2cos4x\left(cos3x+1\right)}=tan4x\)

\(G=\frac{cos2x-sin4x-cos6x}{cos2x+sin4x-cos6x}=\frac{-2sin4xsin2x-sin4x}{-2sin4xsin2x+sin4x}\)

\(G=\frac{-sin4x\left(2sin2x+1\right)}{-sin4x\left(2sin2x-1\right)}=\frac{2sin2x+1}{2sin2x-1}\)

\(P=\dfrac{-2sin5x.sinx-sinx}{2sin5x.cosx+cosx}=\dfrac{-sinx\left(2sin5x+1\right)}{cosx\left(2sin5x+1\right)}=-tanx\)

x=π/2+kπ

x=π/18+kπ/9

a/

\(\Leftrightarrow2sin4x.cos3x=2sin7x.cos3x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\sin7x=sin4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\frac{\pi}{2}+k\pi\\7x=4x+k2\pi\\7x=\pi-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k\pi}{3}\\x=\frac{k2\pi}{3}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

b.

\(\Leftrightarrow2cos4x.cosx=2cos8x.cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=4x+k2\pi\\8x=-4x+k2\pi\end{matrix}\right.\)

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

\(\text{c) }sin3x-\sqrt{3}cos3x=2cos5x\\ \Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=cos5x\\ \Leftrightarrow sin\frac{\pi}{6}\cdot sin3x-cos\frac{\pi}{6}\cdot cos3x=cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=-cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=cos\left(\pi-5x\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=\pi-5x+m2\pi\\3x+\frac{\pi}{6}=5x-\pi+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{48}+\frac{m\pi}{4}\\x=\frac{7\pi}{12}-n\pi\end{matrix}\right.\)

\(d\text{) }sinx\left(sinx+2cosx\right)=2\\ \Leftrightarrow cos^2x+\left(sinx-cosx\right)^2=0\\ \Leftrightarrow cosx=sinx=0\left(VN\right)\)

\(e\text{) }\sqrt{3}\left(sin2x+cos7x\right)=sin7x-cos2x\\ \Leftrightarrow\sqrt{3}sin2x+cos2x=sin7x-\sqrt{3}cos7x\\ \Leftrightarrow sin2x\cdot\frac{\sqrt{3}}{2}+cos2x\cdot\frac{1}{2}=sin7x\cdot\frac{1}{2}-cos7x\cdot\frac{\sqrt{3}}{2}\\ \Leftrightarrow sin2x\cdot cos\frac{\pi}{3}+cos2x\cdot sin\frac{\pi}{3}=sin7x\cdot cos\frac{\pi}{3}-cos7x\cdot sin\frac{\pi}{3}\\ \Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=7x-\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=\frac{4\pi}{3}-7x+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-m2\pi}{5}\\x=\frac{5\pi}{27}+\frac{n2\pi}{9}\end{matrix}\right.\)

\(\text{a) }\sqrt{3}sin2x-cos2x+1=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=-\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2x=\frac{1}{2}\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+m\pi\\x=n\pi\end{matrix}\right.\)

\(\text{b) }pt\Leftrightarrow sin4x=\frac{1-4cosx}{3}\\ \Leftrightarrow sin^24x+cos^24x=\left(\frac{1-cos4x}{3}\right)^2+cos^24x=1\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{arccos\left(-\frac{4}{5}\right)}{4}+\frac{k\pi}{2}\end{matrix}\right.\)

Chọn A.

Ta có:

+ sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x.cos²x = 1 - 2sin²x.cos²x.

+ sin⁴x + cos⁴x = 1 - 3sin²x.cos²x.

Do đó

A = 3(1 - 2sin²x^.cos²x) - 2(1 - 3sin²x.cos²x) = 1.