tan x=2

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

=>sin x và cosx cùng dấu và \(sinx=2\cdot cosx\)

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+4=5\)

=>\(cos^2x=\dfrac{1}{5}\)

=>\(\left[{}\begin{matrix}cosx=\dfrac{1}{\sqrt{5}}\\cosx=-\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)

TH1: \(cosx=\dfrac{1}{\sqrt{5}}\)

=>\(sinx=\sqrt{1-cos^2x}=\dfrac{2}{\sqrt{5}}\)

TH2: cosx=-1/căn 5

=>\(sinx=-\sqrt{1-cos^2x}=-\dfrac{2}{\sqrt{5}}\)

\(Q=\dfrac{sin^3x}{2sinx+cos^3x}\)

\(=\dfrac{\left(2\cdot cosx\right)^3}{2\cdot2cosx+cos^3x}\)

\(=\dfrac{8\cdot cos^3x}{4cosx+cos^3x}=\dfrac{8cos^2x}{4+cos^2x}\)

\(=\dfrac{8\cdot\dfrac{1}{5}}{4+\dfrac{1}{5}}=\dfrac{8}{5}:\dfrac{21}{5}=\dfrac{8}{21}\)

a/

\(\Leftrightarrow cos^3x-sin^3x=cosx+sinx\)

- Với \(cosx=0\Rightarrow sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\) là 1 nghiệm

- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow1-tan^3x=\frac{1}{cos^2x}+tanx.\frac{1}{cos^2x}\)

\(\Leftrightarrow1-tan^3x=1+tan^2x+tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+tan^2x+tanx=0\)

\(\Leftrightarrow tanx\left(2tan^2x+tanx+1\right)=0\)

\(\Leftrightarrow tanx=0\Rightarrow x=k\pi\)

b/

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

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

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

\(\Leftrightarrow cosx-sinx=\left(1+2sinx\right)\left(cosx+sinx\right)\)

\(\Leftrightarrow sinx+sinx.cosx+sin^2x=0\)

\(\Leftrightarrow sinx\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx+cosx=-1\left(1\right)\end{matrix}\right.\)

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

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

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

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

c/

ĐKXĐ: ...

Chia 2 vế cho \(cos^2x\) ta được:

\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)

\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)

\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

a.

\(A=sin^3x\cdot\left(1+\dfrac{cosx}{sinx}\right)+cos^3x\left(1+\dfrac{sinx}{cosx}\right)\)

\(=sin^2x\left(sinx+cosx\right)+cos^2x\left(cosx+sinx\right)\)
=cosx+sinx

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

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

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

a/

\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

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

b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)

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

\(\Leftrightarrow cosx=sin^2x-cos^2x\)

\(\Leftrightarrow cosx=1-2cos^2x\)

\(\Leftrightarrow2cos^2x+cosx-1=0\)

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

\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)

c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)

\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)

\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)

\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)