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: \(cosx\ne0\)

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

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

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

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

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

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

1.

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

Pt trở thành:

\(t^3+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow2t^3+t^2-3=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)

\(\Leftrightarrow t=1\)

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

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

\(\Leftrightarrow...\)

b.

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t^4-3\left(t^2-1\right)-1=0\)

\(\Leftrightarrow t^4-3t^2+2=0\)

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

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

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

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+2\left(sinx+cosx\right)-6sinx.cosx=0\)

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

Pt trở thành:

\(t\left(1-\frac{t^2-1}{2}\right)+2t-3\left(t^2-1\right)=0\)

\(\Leftrightarrow-t^3-6t^2+7t+6=0\)

Nghiệm của pt bậc 3 này rất xấu, chắc bạn ghi ko đúng đề bài

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

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

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.

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

Đặt \(\frac{\sqrt{2}}{\sqrt{3}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sinx.cosa-cosx.sina=cosa\)

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

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

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

b.

\(\frac{1}{2}sin7x+\frac{\sqrt{3}}{2}cos7x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\7x+\frac{\pi}{3}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{84}+\frac{k2\pi}{7}\\x=\frac{5\pi}{84}+\frac{k2\pi}{7}\end{matrix}\right.\)

c.

\(\Leftrightarrow\frac{5}{13}cos2x-\frac{12}{13}sin2x=1\)

Đặt \(\frac{5}{13}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cos2x.cosa-sin2x.sina=1\)

\(\Leftrightarrow cos\left(2x+a\right)=1\)

\(\Leftrightarrow2x+a=k2\pi\)

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

d.

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

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

\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)

e.

\(\Leftrightarrow cosx.cos\left(\frac{\pi}{12}\right)-sinx.sin\left(\frac{\pi}{12}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{12}\right)=\frac{1}{2}\)

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

a/ ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k2\pi\\x\ne-\frac{\pi}{6}+k2\pi\\x\ne\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(1+sinx-2sin^2x\right)\)

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

\(\Leftrightarrow\sqrt{3}sinx-cosx=sin2x+\sqrt{3}cos2x\)

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

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

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(cosx+\sqrt{3}sinx\ne0\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)\ne0\Rightarrow...\)

Đặt \(cosx+\sqrt{3}sinx=2sin\left(x+\frac{\pi}{6}\right)=a\) với \(-2\le a\le2\):

\(a=\frac{3}{a}+1\Leftrightarrow a^2-a-3=0\)

\(\Rightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{13}}{2}>2\left(l\right)\\a=\frac{1-\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow2sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{2}\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{4}=sin\alpha\)

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