a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)

b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx

⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x

⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x

⇔ 4sin2x + (sinx + cosx) . sin2x = 0

⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)

⇔ sin2x = 0

c, 2cos³x = sin3x

⇔ 2cos³x = 3sinx - 4sin³x

⇔ 4sin³x + 2cos³x - 3sinx(sin²x + cos²x) = 0

⇔ sin³x + 2cos³x - 3sinx.cos²x = 0

Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình

Xét cosx ≠ 0 chia cả 2 vế cho cos³x ta được : 

tan³x + 2 - 3tanx = 0

⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)

d, cos²x - \(\sqrt{3}sin2x\) = 1 + sin²x

⇔ cos²x - sin²x - \(\sqrt{3}sin2x\) = 1

⇔ cos2x - \(\sqrt{3}sin2x\) = 1

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

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

e, cos³x + sin³x = 2cos⁵x + 2sin⁵x

⇔ cos³x (1 - 2cos²x) + sin³x (1 - 2sin²x) = 0

⇔ cos³x . (- cos2x) + sin³x . cos2x = 0

⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)

\(sinx+cosx=m\Leftrightarrow\left(sinx+cosx\right)^2=m^2\)

\(\Leftrightarrow1+2sinx.cosx=m^2\Rightarrow sinx.cosx=\dfrac{m^2-1}{2}\)

\(A=sin^2x+cos^2x=1\)

\(B=sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)

\(=m^3-\dfrac{3m\left(m^2-1\right)}{2}=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{3m-m^3}{2}\)

\(C=\left(sin^2+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1-2\left(\dfrac{m^2-1}{2}\right)^2\)

\(D=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3\left(sin^2x+cos^2x\right)\left(sinx.cosx\right)^2\)

\(=1-3\left(\dfrac{m^2-1}{2}\right)^2\)

tan x=2

=>\(\dfrac{sinx}{cosx}=2\)

=>\(sinx=2\cdot cosx\)

\(B=\dfrac{cos^3x+cosx\cdot sin^2x-sin^3x}{sin^3x-cos^3x}\)

\(=\dfrac{cos^3x+cosx\cdot4cos^2x-8cos^3x}{8cos^3x-cos^3x}\)

\(=\dfrac{-3cos^3x}{7cos^3x}=-\dfrac{3}{7}\)

\(\frac{cos^3x+sin^3x}{1-sinx.cosx}-sinx+cosx=\frac{\left(cosx+sinx\right)\left(cos^2x+sin^2x-sinx.cosx\right)}{1-sinx+cosx}-sinx+cosx\)

\(=\frac{\left(cosx+sinx\right)\left(1-sinx.cosx\right)}{1-sinx.cosx}-sinx+cosx\)

\(=cosx+sinx-sinx+cosx=2cosx\)

Vẫn phụ thuộc biến, chắc bạn ghi đề ko đúng (đoán là chỗ \(-sinx+cosx\) có ngoặc chung)

câu này nhìn ngứa mắt quá làm kiểu gì giờ ??? 

a/ \(\Leftrightarrow2cosx.cos2x=cos2x\)

\(\Leftrightarrow2cosx.cos2x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cosx=\frac{1}{2}\end{matrix}\right.\)

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

b/ \(\Leftrightarrow2sinx.sin2x=sinx\)

\(\Leftrightarrow2sinx.sin2x-sinx=0\)

\(\Leftrightarrow sinx\left(2sin2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\sin2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=\frac{\pi}{6}+k2\pi\\2x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

c/ \(\Leftrightarrow sin3x-sinx+sin4x-sin2x=0\)

\(\Leftrightarrow2cos2x.sinx+2cos3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos2x+cos3x\right)=0\)

\(\Leftrightarrow2sinx.2cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{5}+\frac{k4\pi}{5}\\x=\pi+k4\pi\end{matrix}\right.\)

d/ \(\Leftrightarrow sin3x-sinx-\left(sin4x-sin2x\right)=0\)

\(\Leftrightarrow2cos2x.sinx-2cos3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos2x-cos3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=cos3x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=3x+k2\pi\\2x=-3x+k2\pi\end{matrix}\right.\)

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