4.

ĐKXĐ: \(2cos^2x+sinx-1\ne0\)

\(\Leftrightarrow-2sin^2x+sinx+1\ne0\Rightarrow\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)

Khi đó pt tương đương:

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(loại\right)\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

3.

\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)

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

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

Nhiều quá @@ Tách ra đi ><

c/

Đặt \(3cosx-4sinx-6=t\)

Pt trở thành:

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

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

\(\Rightarrow\left[{}\begin{matrix}3cosx-4sinx-6=-1\\3cosx-4sinx-6=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3cosx-4sinx=5\\3cosx-4sinx=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx.\frac{3}{5}-sinx.\frac{4}{5}=1\\cosx.\frac{3}{5}-sinx.\frac{4}{5}=\frac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+a\right)=1\\cosx\left(x+a\right)=\frac{4}{5}\end{matrix}\right.\) (với góc \(a\in\left[0;\pi\right]\) sao cho \(cosa=\frac{3}{5}\))

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

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

a/

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

\(\Leftrightarrow cosx.cos\left(\frac{\pi}{3}\right)-sinx.sin\left(\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}-x\right)\)

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

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

\(\Rightarrow x=k\pi\)

b/

\(\Leftrightarrow\sqrt{2}sin\left(5x+\frac{\pi}{4}\right)=\sqrt{2}cos13x\)

\(\Leftrightarrow cos\left(\frac{\pi}{4}-5x\right)=cos13x\)

\(\Leftrightarrow\left[{}\begin{matrix}13x=\frac{\pi}{4}-5x+k2\pi\\13x=-\frac{\pi}{4}+5x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{72}+\frac{k\pi}{9}\\x=-\frac{\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\dfrac{cosx-2sinx.cosx}{1-2sin^2x+sinx}=\sqrt{3}\)

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

ĐKXĐ : \(sinx\ne1;-\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+2k\pi\\x\ne\dfrac{-\pi}{6}+2k\pi;\dfrac{7\pi}{6}+2k\pi\end{matrix}\right.\)   

\(\Leftrightarrow x\ne\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\)( k thuộc Z ) 

P/t đã cho \(\Leftrightarrow\dfrac{cosx-sin2x}{1-2sin^2x+sinx}=\sqrt{3}\) 

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+2k\pi\\2x+\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+2k\pi\end{matrix}\right.\) ( k thuộc Z ) 

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

Vậy ...

\(\Leftrightarrow2\left(\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx\right)+2cos\left(x-\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=1\)

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

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

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

a: sin 5x=sin x

=>\(\left[{}\begin{matrix}5x=x+k2\Omega\\5x=\Omega-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=k2\Omega\\6x=\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{k\Omega}{2}\\x=\dfrac{\Omega}{6}+\dfrac{k\Omega}{3}\end{matrix}\right.\)

b; \(cos\left(x+\dfrac{\Omega}{3}\right)=-\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=\dfrac{2}{3}\Omega+k2\Omega\\x+\dfrac{\Omega}{3}=-\dfrac{2}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)

ĐKXĐ: ...

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=\dfrac{3}{2}\left(1+tan^2x\right)-\sqrt{3}tanx\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\)

\(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)\le1\\\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\ge1\end{matrix}\right.\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=1\\tanx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)

Hình như có nhầm lẫn từ dòng 1 xuống dòng 2 thì phải. Em bấm máy tính ra nghiệm pi/6 mà.