a, \(cos^2x-cosx=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)

a, \(cos^2x-cosx=0\)

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

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

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

Vậy...

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)

\(\Leftrightarrow8sin^2x-2sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

Vậy...

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\) 

\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)

\(\Leftrightarrow cos2x=\dfrac{1}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)

Vậy...

e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))

\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)

\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow-8sin^2x+2sinx+1=0\)

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

Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0\)

\(\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)\)

\(\Leftrightarrow cos^2x=\dfrac{3}{4}\)

\(\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}\)

Với \(cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)

Với \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)

a/

\(\Leftrightarrow4cos^2\left(6x-2\right)+8\left(1+cos\left(6x-2\right)\right)-13=0\)

Đặt \(cos\left(6x-2\right)=a\Rightarrow\left|a\right|\le1\)

Pt trở thành:

\(4a^2+8\left(1+a\right)-13=0\)

\(\Leftrightarrow4a^2+8a-5=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{2}\\a=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos\left(6x-2\right)=\frac{1}{2}\)

\(\Rightarrow6x-2=\pm\frac{\pi}{3}+k2\pi\)

\(\Rightarrow x=\frac{1}{3}\pm\frac{\pi}{18}+\frac{k\pi}{3}\)

b/

\(\Leftrightarrow2cos^2\left(x+75^0\right)-1+3sin\left(15^0-x\right)-1=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(90^0-15^0+x\right)-2=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(x+75^0\right)-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(x+75^0\right)=\frac{1}{2}\\cos\left(x+75^0\right)=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+75^0=60^0+k360^0\\x+75^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+k360^0\\x=-135^0+k360^0\end{matrix}\right.\)

c/

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

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow cos2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow1-2sin^2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)\left(1-2sin\left(x-\frac{\pi}{6}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{6}\right)=0\\sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=k\pi\\x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow\left(sinx-cosx\right)^2+2\left|sinx-cosx\right|\left|sinx+cosx\right|+\left(cosx+sinx\right)^2=4\)

\(\Leftrightarrow2\left(sin^2x+cos^2x\right)+2\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|=4\)

\(\Leftrightarrow\left|sin^2x-cos^2x\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=1\\sin^2x-cos^2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=sin^2x+cos^2x\\sin^2x-cos^2x=-\left(sin^2x+cos^2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=0\\sin^2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=0\end{matrix}\right.\)\(\Rightarrow cosx.sinx=0\Rightarrow sin2x=0\)

\(\Rightarrow x=\dfrac{k\pi}{2},k\in Z\)

Vậy...

b) ĐK:\(x\ne\dfrac{k\pi}{2};k\in Z\)

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

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{cosx.sinx}=4\left(sinx+\sqrt{3}cosx\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\left(1\right)\\\dfrac{sinx-\sqrt{3}cosx}{sinx.cosx}=4\left(2\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow tanx=-\sqrt{3}\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi,k\in Z\)

Từ (2)\(\Leftrightarrow sinx-\sqrt{3}cosx=4sinx.cosx\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=2sinx.cosx\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin2x\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)

c) ĐK: \(x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)

Pt \(\Leftrightarrow\left(\sqrt{2}sinx-1\right)^2+\left(\sqrt{3}tan2x-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}sinx-1=0\\\sqrt{3}tan2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sinx=\dfrac{1}{\sqrt{2}}\\tan2x=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm

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/

\(\Leftrightarrow tan2x=-tan40^0\)

\(\Leftrightarrow tan2x=tan\left(-40^0\right)\)

\(\Rightarrow2x=-40^0+k180^0\)

\(\Rightarrow x=-20^0+k90^0\)

b/

\(\Leftrightarrow tan\left(2x-15^0\right)=1\)

\(\Rightarrow2x-15^0=45^0+k180^0\)

\(\Rightarrow x=30^0+k90^0\)

c/

\(\Leftrightarrow tan\left(60^0-x\right)=-\frac{1}{\sqrt{3}}\)

\(\Rightarrow60^0-x=-30^0+k180^0\)

\(\Rightarrow x=90^0+k180^0\)

d/

\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=-tan\left(\frac{\pi}{5}\right)\)

\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=tan\left(-\frac{\pi}{5}\right)\)

\(\Rightarrow3x+\frac{2\pi}{5}=-\frac{\pi}{5}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{5}+\frac{k\pi}{3}\)

a/

\(\Leftrightarrow tanx=-tan\left(\frac{2\pi}{3}-3x\right)\)

\(\Leftrightarrow tanx=tan\left(3x-\frac{2\pi}{3}\right)\)

\(\Rightarrow x=3x-\frac{2\pi}{3}+k\pi\)

\(\Rightarrow x=\frac{\pi}{3}+\frac{k\pi}{2}\)

b/

\(tan\left(2x-15^0\right)=tanx\)

\(\Rightarrow2x-15^0=x+k180^0\)

\(\Rightarrow x=15^0+k180^0\)

c/

ĐKXĐ: ...

\(\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)\)

\(\Leftrightarrow5tan2x=-5\)

\(\Rightarrow tan2x=-1\)

\(\Rightarrow2x=-\frac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)

d/

ĐKXĐ: ...

\(\Leftrightarrow sinx+\sqrt{3}cosx=3sinx-\sqrt{3}cosx\)

\(\Leftrightarrow2sinx=2\sqrt{3}cosx\)

\(\Rightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)

a/

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

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

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

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

b/ ĐKXĐ: ...

\(\Leftrightarrow tanx+\frac{1}{tanx}=2\)

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

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

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)

c/

\(a+b+c=1+\sqrt{3}-1-\sqrt{3}=0\)

\(\Rightarrow\) Pt có 2 nghiệm: \(\left[{}\begin{matrix}tanx=1\\tanx=-\sqrt{3}\end{matrix}\right.\)

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

d/ ĐKXĐ: ...

\(\Leftrightarrow cot^22x+3.cot2x+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cot2x=-1\\cot2x=-2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{1}{2}arccot\left(-2\right)+\frac{k\pi}{2}\end{matrix}\right.\)