a/

\(\Leftrightarrow2cos2x.cosx+\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos^22x=0\)

\(\Leftrightarrow cos2x\left(2cosx+cos2x\right)=0\)

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

\(\left(1\right)\Leftrightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(\left(2\right)\Leftrightarrow2cosx+2cos^2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}-1}{2}\\cosx=\frac{-\sqrt{3}-1}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\)

b/

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

\(\Leftrightarrow cos^2x+\frac{1}{2}cosx=0\)

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

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

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

c/ ĐKXĐ: ...

\(\Leftrightarrow\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{sinx.cosx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{2}{sin2x}\right)^2+\frac{3}{sin2x}-7=0\)

Đặt \(\frac{1}{sin2x}=a\Rightarrow4a^2+3a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=arcsin\left(-\frac{4}{7}\right)+k2\pi\\2x=\pi-arcsin\left(-\frac{4}{7}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\end{matrix}\right.\)

a.

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

b.

\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)

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

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

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

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

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

c.

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

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

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

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

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

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

\(\Leftrightarrow...\)

a/

\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x-4cosx=0\)

\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx-\sqrt{3}cosx=2\left(1\right)\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow x=\frac{5\pi}{6}+k2\pi\)

b/

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

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

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

\(\left(1\right)\Leftrightarrow tanx=\sqrt{3}\)

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

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

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

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

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

c/

\(\Leftrightarrow sin6x\left(cos3x-1-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\Rightarrow x=\frac{k\pi}{6}\\cos3x-sin3x=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin3x-cos3x=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\3x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

Bạn tham khảo pt 1 hộ mình nha. Chúc bạn học tốt~

Pt 1.

Bạn tham khảo phương trình 1 hộ mình nha. Chúc bạn học tốt

\(sin4x-2cos2x.cosx=0\)

\(\Leftrightarrow2sin2x.cos2x-2cos2x.cosx=0\)

\(\Leftrightarrow cos2x\left(sin2x-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin2x-cosx=0\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow sin2x=cosx=sin\left(\frac{\pi}{2}-x\right)\)

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

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

\(\left(cosx+sin2x\right).sin2x=0\)

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

\(\Leftrightarrow\left(1+2sinx\right)sinx.cos^2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\2x=k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\end{matrix}\right.\)

\(cosx+cos3x+cos2x+cos4x=0\)

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

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

\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)

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

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

a/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+sin2x-1=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+sin2x-1=0\)

\(\Leftrightarrow-\frac{3}{4}sin^22x+sin2x=0\)

\(\Leftrightarrow sin2x\left(1-\frac{3}{4}sin2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=k\pi\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

b/

\(\Leftrightarrow\left(1+sin2x\right)+sinx+cosx+cos^2x-sin^2x=0\)

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

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

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

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

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

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

c/

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=0\\sin\left(x-\frac{\pi}{3}\right)=1\end{matrix}\right.\)

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

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

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

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

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)