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.

a. \(sin\left(4x+\pi\right)=sin35^o\)

\(\Leftrightarrow sin\left(4x+180^o\right)=sin35^o\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+180^o=35^o+k.360^o,k\in Z\\4x+180^o=180^o-35^o+k.360^o,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-145^o+k.360^o,k\in Z\\4x=-35^o+k.360^o,k\in Z\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{145^o}{4}+k.90,k\in Z\\x=-\frac{35^o}{4}+k.90^o,k\in Z\end{matrix}\right.\)

Vậy.....

b.\(sin4x=\frac{1}{5}\)

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

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

Vậy....

c. \(sin\left(x+\frac{8\pi}{7}\right)=3\)

Ta có: \(-1\le sinx\le1\)

\(\Rightarrow-1\le sin\left(3x+\frac{8\pi}{7}\right)\le1\)

Do đó phương trình trên vô nghiệm

d. \(sinx=-7\)

Ta có: \(-1\le sinx\le1\)

Do đó phương trình trên vô nghiệm

e. \(sin\left(3x+\pi\right)=sin\left(2x-3\pi\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\5x=3\pi+k2\pi,k\in Z\end{matrix}\right.\)

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

Vậy......

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

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

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

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

Vậy......

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

\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k2\pi\)

b.

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

\(\Leftrightarrow cos2x-sin2x+sinx+cosx=1\)

\(\Leftrightarrow1-2sin^2x-2sinx.cosx+sinx+cosx=1\)

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

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

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

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.

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

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

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

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

\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)

\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)

\(=sinx+cosx-cosx=sinx\)

\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)

\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)

\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)

\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)

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

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)

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

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

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

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

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

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

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