a)

B1

b)

@Nguyễn Việt Lâm giúp em với ạ

a) Với mọi \(x \in \mathbb{R}\) ta có \( - 1 \le cosx \le 1\)

Vậy phương trình \(cosx =  - 3\;\) vô nghiệm.

\(\begin{array}{l}b)\,\;cosx = cos{15^o}\;\\ \Leftrightarrow \left[ \begin{array}{l}x = {15^o} + k{360^o},k \in \mathbb{Z}\\x =  - {15^o} + k{360^o},k \in \mathbb{Z}\end{array} \right.\end{array}\)

Vậy phương trình có nghiệm \(x = {15^o} + k{360^o}\) hoặc \(x =  - {15^o} + k{360^o},k \in \mathbb{Z}\).

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

Vậy phương trình có nghiệm \(x = \frac{\pi }{6} + k2\pi ,\) hoặc \(x =  - \frac{\pi }{3} + k2\pi ,k \in \mathbb{Z}\).

a, Điều kiện xác định: \(\frac{1}{2}x + \frac{\pi }{4} \ne k\pi  \Leftrightarrow x \ne  - \frac{\pi }{2} + k2\pi ,k \in \mathbb{Z}.\)

Ta có: \(cot\left( {\frac{1}{2}x + \frac{\pi }{4}} \right) =  - 1 \Leftrightarrow cot\left( {\frac{1}{2}x + \frac{\pi }{4}} \right) = \cot \left( { - \frac{\pi }{4}} \right)\)

\( \Leftrightarrow \frac{1}{2}x + \frac{\pi }{4} =  - \frac{\pi }{4} + k\pi  \Leftrightarrow x =  - \pi  + k2\pi ,k \in \mathbb{Z}\,\,(TM).\)

Vậy \(x =  - \pi  + k2\pi ,k \in \mathbb{Z}\,\).

b, Điều kiện xác định: \(3x \ne k\pi  \Leftrightarrow x \ne k\frac{\pi }{3},k \in \mathbb{Z}.\)

\(\;cot3x =  - \frac{{\sqrt 3 }}{3} \Leftrightarrow cot3x = \cot \left( { - \frac{\pi }{3}} \right)\)

\( \Leftrightarrow 3x =  - \frac{\pi }{3} + k\pi  \Leftrightarrow x =  - \frac{\pi }{9} + k\frac{\pi }{3},k \in \mathbb{Z}\,\,(TM).\)

Vậy \(x =  - \frac{\pi }{9} + k\frac{\pi }{3},k \in \mathbb{Z}\,\).

1.

\(\Leftrightarrow\frac{\pi}{3}cosx-\frac{8\pi}{3}=k\pi\)

\(\Leftrightarrow cosx=8+3k\)

Do \(-1\le cosx\le1\Rightarrow-1\le8+3k\le1\)

\(\Rightarrow-3\le k\le-\frac{7}{3}\) \(\Rightarrow k=-3\)

\(\Rightarrow cosx=-1\Rightarrow x=\pi+k2\pi\)

2.

\(\Leftrightarrow\frac{\pi}{3}cos2\pi x=\frac{\pi}{6}+k\pi\)

\(\Leftrightarrow cos2\pi x=\frac{1}{2}+3k\)

Do \(-1\le2\pi x\le1\Rightarrow-1\le\frac{1}{2}+3k\le1\)

\(\Rightarrow-\frac{1}{2}\le k\le\frac{1}{6}\Rightarrow k=0\)

\(\Rightarrow cos2\pi x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2\pi x=\frac{\pi}{3}+k2\pi\\2\pi x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}+k\\x=-\frac{1}{6}+k\end{matrix}\right.\)

@Nguyễn Việt Lâm giúp em với ạ

1.

ĐKXĐ: \(cos\left(4x+\frac{2\pi}{5}\right)+cos\left(3x-\frac{\pi}{4}\right)\ne0\)

\(\Leftrightarrow cos\left(4x+\frac{2\pi}{5}\right)\ne cos\left(3x+\frac{3\pi}{4}\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+\frac{2\pi}{5}\ne3x+\frac{3\pi}{4}+k2\pi\\4x+\frac{2\pi}{5}\ne-3x-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{7\pi}{20}+k2\pi\\x\ne-\frac{23\pi}{140}+\frac{k2\pi}{7}\end{matrix}\right.\)

2.

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\left(\frac{\sqrt{2}}{3}\right)+k4\pi\\x=-2arccos\left(\frac{\sqrt{2}}{3}\right)+k4\pi\end{matrix}\right.\)

\(\Rightarrow x=4\pi-2arccos\left(\frac{\sqrt{2}}{3}\right)\approx10.41\left(rad\right)\)

3.

a.

\(\Leftrightarrow\left(cos3x-cosx\right)+\left(cos2x-1\right)=0\)

\(\Leftrightarrow-2sin2x.sinx+1-2sin^2x-1=0\)

\(\Leftrightarrow sin2x.sinx+sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx+sin^2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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