a/

\(\Leftrightarrow sinx+cosx=\sqrt{2}sin2x\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}sin2x\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=sin2x\)

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

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

b/

\(\Leftrightarrow\frac{1-cos2x}{2}+sin2x=\frac{3\left(1+cos2x\right)}{2}\)

\(\Leftrightarrow sin2x-2cos2x=1\)

\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\frac{1}{\sqrt{5}}\)

Đặt \(\frac{1}{\sqrt{5}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Leftrightarrow sin2x.cosa-cos2a.sina=cosa\)

\(\Leftrightarrow sin\left(2x-a\right)=cosa=sin\left(\frac{\pi}{2}-a\right)\)

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

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

c/

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

\(\Leftrightarrow sinx-cosx-\left(sin^2x-cos^2x\right)=0\)

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

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

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

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

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

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

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

d/

\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)

\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)

\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)

\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)

a,

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

b,

\(2sinx+2\sqrt{3}cosx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\\ < =>2sinx-\dfrac{1}{sinx}=\dfrac{\sqrt{3}}{cosx}-2\sqrt{3}cosx\\ < =>\dfrac{2sin^2x-1}{sinx}=\dfrac{\sqrt{3}.cosx.\left(1-2cos^2x\right)}{cosx}\\ < =>\dfrac{cos2x}{sinx}=\sqrt{3}.cos2x\\ < =>\dfrac{1}{sinx}=\sqrt{3}\)

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

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

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

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

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

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

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

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

...

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: \(x\ne\frac{\pi}{2}+k\pi\)

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

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

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

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

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

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

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

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

a/

\(\Leftrightarrow1-2\left(2cos^2x-1\right)-\sqrt{3}sinx+cosx=0\)

\(\Leftrightarrow3-4cos^2x+cosx-\sqrt{3}sinx=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(4cosx+3\right)-\sqrt{3}sinx=0\)

\(\Leftrightarrow2sin^2\frac{x}{2}\left(4cosx+3\right)-2\sqrt{3}sin\frac{x}{2}cos\frac{x}{2}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}sin\frac{x}{2}=0\Rightarrow x=k2\pi\\sin\frac{x}{2}\left(4cosx+3\right)-\sqrt{3}cos\frac{x}{2}=0\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow sin\frac{x}{2}\left(8cos^2\frac{x}{2}-1\right)-\sqrt{3}cos\frac{x}{2}=0\)

- Với \(\left\{{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-1\end{matrix}\right.\) \(\Rightarrow x=-\pi+k4\pi\) là 1 nghiệm

- Với \(cos\frac{x}{2}\ne0\) chia 2 vế cho \(cos^3\frac{x}{2}\)

\(tan\frac{x}{2}\left(8-1-tan^2\frac{x}{2}\right)-\sqrt{3}-\sqrt{3}tan^2\frac{x}{2}=0\)

\(\Leftrightarrow-tan^3\frac{x}{2}-\sqrt{3}tan^2\frac{x}{2}+7tan\frac{x}{2}-\sqrt{3}=0\)

Đặt \(tan\frac{x}{2}=t\)

\(\Rightarrow t^3+\sqrt{3}t^2-7t+\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\sqrt{3}\\t=-2-\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{3}+k\pi\\\frac{x}{2}=-\frac{5\pi}{12}+k\pi\\\frac{x}{2}=\frac{\pi}{12}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{5\pi}{6}+k2\pi\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow cos^2x-sin^2x+cos^2x-sinx.cosx=8\left(cosx-sinx\right)\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)+cosx\left(cosx-sinx\right)=8\left(cosx-sinx\right)\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(2cosx+sinx-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\left(1\right)\\2cosx+sinx=8\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x-\frac{\pi}{4}=k\pi\)

\(\Rightarrow x=\frac{\pi}{4}+k\pi\)

Xét (2), theo điều kiện có nghiệm của pt lượng giác bậc nhất, \(2^2+1^2< 8^2\Rightarrow\left(2\right)\) vô nghiệm

c/

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx\right)=4\left(sinx-cosx\right)\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\sinx+4cosx-4=0\left(2\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

Xét (2) \(\Leftrightarrow\frac{1}{\sqrt{17}}sinx+\frac{4}{\sqrt{17}}cosx=\frac{4}{\sqrt{17}}\)

Đặt \(\frac{4}{\sqrt{17}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cosx.cosa+sinx.sina=cosa\)

\(\Leftrightarrow cos\left(x-a\right)=cosa\)

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

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