a) \(\cos \left( {3x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} =  - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi  + k2\pi }\\{3x =  - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x =  - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} =  - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} =  - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x =  - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi  + k2\pi }\\{x =  - \pi  + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)

c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x =  - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

Câu 2 bạn coi lại đề

3.

\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)

\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)

\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)

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

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

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

\(\Leftrightarrow...\)

4.

Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm

5.

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)

\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)

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

Xét (1):

\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)

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

\(\Leftrightarrow2sin^3x-sinx-1=0\)

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

\(\Leftrightarrow...\)

6.

\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)

\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)

\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a.

\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)

c.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)

d.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)

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

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)

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

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} =  - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x =  - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)     \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} =  - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} =  - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} =  - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{6} + k4\pi \\x =  - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

d)     \(2\cos 3x + 5 = 3\)

\(\begin{array}{l} \Leftrightarrow \cos 3x =  - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi  + k2\pi \\3x =  - \pi  + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

e)      

\(\begin{array}{l}3\tan x =  - \sqrt 3 \\ \Leftrightarrow \tan x = \frac{{ - \sqrt 3 }}{3}\\ \Leftrightarrow \tan x = \tan \left( { - \frac{\pi }{6}} \right)\\ \Leftrightarrow x =  - \frac{\pi }{6} + k\pi \end{array}\)

g)

\(\begin{array}{l}\cot x - 3 = \sqrt 3 \left( {1 - \cot x} \right)\\ \Leftrightarrow \cot x - 3 = \sqrt 3  - \sqrt 3 \cot x\\ \Leftrightarrow \cot x + \sqrt 3 \cot x = \sqrt 3  + 3\\ \Leftrightarrow (1 + \sqrt 3 )\cot x = \sqrt 3  + 3\\ \Leftrightarrow \cot x = \sqrt 3 \\ \Leftrightarrow \cot x = \cot \frac{\pi }{6}\\ \Leftrightarrow x = \frac{\pi }{6} + k\pi \end{array}\)

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.

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

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

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

\(\Leftrightarrow...\)

\(\text{1) }cos^2\left(x-\frac{\pi}{6}\right)-sin^2\left(x-\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+m2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{m2\pi}{3}\\x=\frac{\pi}{6}+n2\pi\end{matrix}\right.\\\Leftrightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3} \)

\(2\text{) }sin^4x-sin^4\left(x+\frac{\pi}{2}\right)=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow sin^2x-cos^2x=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow cos\left(\pi-2x\right)=cos\left(\frac{\pi}{6}-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}\pi-2x=\frac{\pi}{6}-x+m2\pi\\\pi-2x=x-\frac{\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}-m2\pi\\x=\frac{7\pi}{18}-\frac{n2\pi}{3}\end{matrix}\right.\)

\(3\text{) }pt\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+m2\pi\\x=n2\pi\end{matrix}\right.\)

a/

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

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

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

b/

\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)

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

\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

c/

\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)

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

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

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

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

Ta có:

\({\cos ^2}a + {\sin ^2}a = 1 \Rightarrow \sin a =  \pm \frac{4}{5}\)

Do \(0 < a < \frac{\pi }{2} \Leftrightarrow \sin a = \frac{4}{5}\)

\(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{4}{3}\)

Ta có;

\(\begin{array}{l}\sin \left( {a + \frac{\pi }{6}} \right) = \sin a.\cos \frac{\pi }{6} + \cos a.\sin \frac{\pi }{6} = \frac{4}{5}.\frac{{\sqrt 3 }}{2} + \frac{3}{5}.\frac{1}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\cos \left( {a - \frac{\pi }{3}} \right) = \cos a.\cos \frac{\pi }{3} + \sin a.\sin \frac{\pi }{3} = \frac{3}{5}.\frac{1}{2} + \frac{4}{5}.\frac{{\sqrt 3 }}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\tan \left( {a + \frac{\pi }{4}} \right) = \frac{{\tan a + \tan \frac{\pi }{4}}}{{1 - \tan a.tan\frac{\pi }{4}}} = \frac{{\frac{4}{3} + 1}}{{1 - \frac{4}{3}}} =  - 7\end{array}\)

\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)

\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) =  - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B =  - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) =  - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) =  - \frac{3}{8}\end{array}\)