a/

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow3tanx-\sqrt{3}=0\)

\(\Rightarrow tanx=\frac{1}{\sqrt{3}}\)

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

b/

ĐKXĐ: \(sinx\ne-1\)

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

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

\(\Leftrightarrow\frac{\sqrt{3}cosx-1}{1+sinx}=-1\) (do 2-sinx>0 với mọi x)

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

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

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

c/

ĐKXĐ: \(sin2x\ne0\)

\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)

\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)

\(\Leftrightarrow1-cosx=sin^2x\)

\(\Leftrightarrow1-cosx=1-cos^2x\)

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

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

d/

ĐKXĐ: \(cosx\ne0\)

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

\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)

\(\Leftrightarrow tanx=\sqrt{3}\)

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

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: ...

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

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

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

\( a){\mathop{\rm sinx}\nolimits} + \cos x = \sqrt 2 \sin 5x\\ \Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .\sin 5x\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 5x\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = 5x + k2\pi \\ x + \dfrac{\pi }{4} = \pi - 5x + k2\pi \end{array} \right.\left( {k \in \mathbb {Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)

\( b)\sqrt 3 \sin 2x + \sin \left( {\dfrac{\pi }{2} + 2x} \right) = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \sin \dfrac{\pi }{2}\cos 2x + \cos \dfrac{\pi }{2}\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + 1.\cos 2x + 0.\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \cos 2x - 1 = 0\\ \Leftrightarrow 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} + 1 - 2{\sin ^2}x - 1 = 0\\ \Leftrightarrow \sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - si{n^2}x = 0\\ \Leftrightarrow {\mathop{\rm sinx}\nolimits} \left( {\sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\mathop{\rm sinx}\nolimits} = 0\\ \sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \sin \left( {\dfrac{\pi }{3} - x} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \dfrac{\pi }{3} - x = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{3} - k\pi \end{array} \right. \)

Nhiều quá @@ Tách ra đi ><

\( d)3{\sin ^2}x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow 2{\sin ^2}x + 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - 3 = 0\\ *sinx = 0 \Rightarrow \text{không là nghiệm phương trình}\\ *sin \ne 0\\ 2 + 2\sqrt 3 \cot x - 3\left( {1 + {{\cot }^2}x} \right) = 0\\ \Leftrightarrow 3{\cot ^2}x - 2\sqrt 3 \cot x + 1 = 0\\ \Leftrightarrow \cot x = \dfrac{{\sqrt 3 }}{3} \Rightarrow x = \dfrac{\pi }{3} + k\pi \)

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:

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

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

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

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

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

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

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