1.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cotx-tanx=sinx+cosx\)

\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)

\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=sinx+cosx\)

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

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

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

\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)

\(\Leftrightarrow t^2+2t-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)

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

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

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

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

Vậy phương trình đã cho có nghiệm:

\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)

a: \(\frac{1}{\sin x}+\frac{1}{cosx}=4\cdot\sin\left(x+\frac{\pi}{4}\right)\)

=>\(\frac{\sin x+cosx}{\sin x\cdot cosx}=4\cdot\frac{\sqrt2}{2}\cdot\left(\sin x+cosx\right)\)

=>\(\left(\sin x+cosx\right)\left(\frac{1}{\sin x\cdot cosx}-2\sqrt2\right)=0\)

TH1: \(\frac{1}{\sin x\cdot cosx}-2\sqrt2=0\)

=>\(\frac{1}{\sin x\cdot cosx}=2\sqrt2\)

=>\(sinx\cdot cosx=\frac{1}{2\sqrt2}\)

=>\(2\cdot\sin x\cdot cosx=\frac{1}{\sqrt2}\)

=>\(\sin2x=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{4}+k2\pi\\ 2x=-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{8}+k\pi\\ x=-\frac{\pi}{8}+k\pi\end{array}\right.\)

TH2: sin x+cosx=0

=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)

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

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

a: ĐKXĐ: \(\sqrt3\cdot\sin x+cosx<>0\)

=>\(\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx<>0\)

=>\(\sin\left(x+\frac{\pi}{6}\right)<>0\)

=>\(x+\frac{\pi}{6}<>k\pi\)

=>\(x<>-\frac{\pi}{6}+k\pi\)

\(\frac{2\cdot cos2x+1}{\sqrt3\cdot\sin x+cosx}=2\cdot cosx-1\)

=>\(\frac{2\cdot\left(2\cdot cos^2x-1\right)+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}=2\cdot cosx-1\)

=>\(\frac{4\cdot cos^2x-1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-\left(2\cdot cosx-1\right)=0\)

=>\(\left(2\cdot cosx-1\right)\left\lbrack\frac{2\cdot cosx+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-1\right\rbrack=0\)

TH1: \(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}-1=0\)

=>\(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}=1\)

=>\(\sqrt3\cdot\sin x+cosx=2\cdot cosx+1\)

=>\(\sqrt3\cdot\sin x-cosx=1\)

=>\(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx=\frac12\)

=>\(\sin\left(x-\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ x-\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=\pi+k2\pi\end{array}\right.\)

TH2: \(2\cdot cosx-1=0\)

=>\(cosx=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=-\frac{\pi}{3}+k2\pi\end{array}\right.\)

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.

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

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

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

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

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

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

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

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

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

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,

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

f: \(cos7x-\sqrt3\cdot\sin7x-\sin x=\sqrt3\cdot cosx\)

=>\(\frac12\cdot cos7x-\frac{\sqrt3}{2}\cdot\sin7x=\frac12\cdot\sin x+\frac{\sqrt3}{2}\cdot cosx\)

=>\(\sin\left(\frac{\pi}{6}-7x\right)=\sin\left(x+\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}-7x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\ -7x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi=-x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}-7x-x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ -7x+x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-8x=\frac{\pi}{6}+k2\pi\\ -6x=\frac12\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{48}-\frac{k\pi}{4}\\ x=-\frac{1}{12}\pi-\frac{k\pi}{3}\end{array}\right.\)

e: \(5\cdot\sin2x-6\cdot cos^2x=13\)

=>\(5\cdot\sin2x-6\cdot\frac{1+cos2x}{2}=13\)

=>\(5\cdot\sin2x-3-3\cdot cos2x=13\)

=>\(5\cdot\sin2x-3\cdot cos2x=16\)

Vì \(5^2+\left(-3\right)^2=25+9=34<16^2\)

nên phương trình vô nghiệm