a.

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

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)

\(tan3x-tanx=0\)

\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)

\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow2sinx.cosx=0\)

\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)

c.

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)

\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)

\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

d.

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

\(\Leftrightarrow cos\left(2x-1\right)=0\)

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)

e.

\(cos3x+cosx+cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)

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

\(\Leftrightarrow...\)

f.

\(\Leftrightarrow4sin4x.cos4x=\sqrt{2}\)

\(\Leftrightarrow2sin8x=\sqrt{2}\)

\(\Leftrightarrow sin8x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

\(cos5x.cos3x+sin7x.sinx=\frac{1}{2}cos8x+\frac{1}{2}cos2x-\frac{1}{2}cos8x+\frac{1}{2}cos6x\)

\(=\frac{1}{2}\left(cos6x+cos2x\right)=cos4x.cos2x\)

\(\frac{1-2sin^22x}{1-sin4x}=\frac{cos^22x-sin^22x}{cos^22x+sin^22x-2sin2x.cos2x}\)

\(=\frac{\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)}{\left(cos2x-sin2x\right)^2}=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{\frac{cos2x}{cos2x}+\frac{sin2x}{cos2x}}{\frac{cos2x}{cos2x}-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)

\(2cosx-3cos\left(\pi-x\right)+5sin\left(4\pi-\frac{\pi}{2}-x\right)+cot\left(\pi+\frac{\pi}{2}-x\right)\)

\(=2cosx+3cosx-5sin\left(\frac{\pi}{2}+x\right)+cot\left(\frac{\pi}{2}-x\right)\)

\(=5cosx-5cosx+tanx=tanx\)

a/

\(\Leftrightarrow cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}sin\left(\frac{3x}{2}+\frac{\pi}{10}\right)\)

Đặt \(\frac{x}{2}+\frac{\pi}{5}=a\Rightarrow\frac{x}{2}=a-\frac{\pi}{5}\Rightarrow\frac{3x}{2}=3a-\frac{3\pi}{5}\)

Pt trở thành:

\(cosa=\frac{1}{2}sin\left(3a-\frac{3\pi}{5}+\frac{\pi}{10}\right)\)

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

\(\Leftrightarrow cosa=-\frac{1}{2}sin\left(\frac{\pi}{2}-3a\right)=-\frac{1}{2}cos3a\)

\(\Leftrightarrow cos3a+2cosa=0\)

\(\Leftrightarrow4cos^3a-3cosa+2cosa=0\)

\(\Leftrightarrow4cos^3a-cosa=0\)

\(\Leftrightarrow cosa\left(4cos^2a-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosa=0\\cosa=\frac{1}{2}\\cosa=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=0\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}+\frac{\pi}{5}=\frac{\pi}{2}+k\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{\pi}{3}+k2\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\) (5 nghiệm bạn tự biến đổi)

b/

ĐKXĐ: ...

Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)

Pt trở thành:

\(4\left(a^2-2\right)+4a=7\)

\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)

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

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

c/

ĐKXĐ: ...

Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)

Pt trở thành:

\(9a+2\left(a^2-4\right)=1\)

\(\Leftrightarrow2a^2+9a-9=0\)

Pt này nghiệm xấu quá bạn :(

d/ĐKXĐ: ...

Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)

Pt trở thành:

\(2\left(a^2+4\right)+9a-1=0\)

\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

\(A=cos\left(6\pi+\pi-x\right)+sin\left(2\pi+\frac{\pi}{2}-x\right)+tan^2\left(\pi+\frac{\pi}{2}-x\right)-\frac{1}{sin^2\left(7\pi+\pi+x\right)}\)

\(=cos\left(\pi-x\right)+sin\left(\frac{\pi}{2}-x\right)+tan^2\left(\frac{\pi}{2}-x\right)-\frac{1}{sin^2\left(\pi+x\right)}\)

\(=-cosx+cosx+cot^2x-\frac{1}{sin^2x}\)

\(=cot^2x-\left(1+cot^2x\right)=-1\)

@Nguyễn Việt Lâm giúp em với ạ

1.

ĐKXĐ: \(cos\left(4x+\frac{2\pi}{5}\right)+cos\left(3x-\frac{\pi}{4}\right)\ne0\)

\(\Leftrightarrow cos\left(4x+\frac{2\pi}{5}\right)\ne cos\left(3x+\frac{3\pi}{4}\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+\frac{2\pi}{5}\ne3x+\frac{3\pi}{4}+k2\pi\\4x+\frac{2\pi}{5}\ne-3x-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{7\pi}{20}+k2\pi\\x\ne-\frac{23\pi}{140}+\frac{k2\pi}{7}\end{matrix}\right.\)

2.

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\left(\frac{\sqrt{2}}{3}\right)+k4\pi\\x=-2arccos\left(\frac{\sqrt{2}}{3}\right)+k4\pi\end{matrix}\right.\)

\(\Rightarrow x=4\pi-2arccos\left(\frac{\sqrt{2}}{3}\right)\approx10.41\left(rad\right)\)

3.

a.

\(\Leftrightarrow\left(cos3x-cosx\right)+\left(cos2x-1\right)=0\)

\(\Leftrightarrow-2sin2x.sinx+1-2sin^2x-1=0\)

\(\Leftrightarrow sin2x.sinx+sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx+sin^2x=0\)

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

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

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

\(sin\left(\frac{\pi}{7}\right)H=sin\left(\frac{\pi}{7}\right)cos\left(\frac{2\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\)

\(=\frac{1}{2}\left[sin\left(\frac{3\pi}{7}\right)-sin\left(\frac{\pi}{7}\right)+sin\left(\frac{5\pi}{7}\right)-sin\left(\frac{3\pi}{7}\right)+sin\pi-sin\left(\frac{5\pi}{7}\right)\right]\)

\(=-\frac{1}{2}sin\left(\frac{\pi}{7}\right)\)

\(\Rightarrow H=-\frac{1}{2}\)

\(sinA+sinB+sinC=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+2sin\left(\frac{C}{2}\right)cos\left(\frac{C}{2}\right)\)

\(=2cos\frac{C}{2}cos\left(\frac{A-B}{2}\right)+2cos\left(\frac{A+B}{2}\right)cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left[cos\left(\frac{A-B}{2}\right)+cos\left(\frac{A+B}{2}\right)\right]\)

\(=4cos\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}\)

\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)

\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)

\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)

\(=\frac{1}{4}sin3a\)

\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)

\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)

\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)

\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)

\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)

@Nguyễn Việt Lâm giúp em với ạ

a.

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

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

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

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

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

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

b.

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

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

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

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

Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)

\(\Rightarrow\frac{1-t^2}{2}+t=0\)

\(\Leftrightarrow-t^2+2t+1=0\Rightarrow\left[{}\begin{matrix}t=1-\sqrt{2}\\t=1+\sqrt{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)

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

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

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