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giải PT:

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Nguyễn Việt Lâm
3 tháng 8 2024 lúc 21:13

a.

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{5\pi}{6}\right)\)

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

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

Nguyễn Lê Phước Thịnh
3 tháng 8 2024 lúc 21:15

 

a: \(2\cdot cos\left(x+\dfrac{\Omega}{6}\right)=-\sqrt{3}\)

=>\(cos\left(x+\dfrac{\Omega}{6}\right)=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{6}=\dfrac{5}{6}\Omega+k2\Omega\\x+\dfrac{\Omega}{6}=-\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)

b: \(sin\left(2x+\dfrac{\Omega}{4}\right)+cosx=0\)

=>\(sin\left(2x+\dfrac{\Omega}{4}\right)=-cosx=cos\left(\Omega-x\right)\)

=>\(sin\left(2x+\dfrac{\Omega}{4}\right)=sin\left(\dfrac{\Omega}{2}-\Omega+x\right)=sin\left(x-\dfrac{\Omega}{2}\right)\)

=>\(\left[{}\begin{matrix}2x+\dfrac{\Omega}{4}=x-\dfrac{\Omega}{2}+k2\Omega\\2x=\Omega-x+\dfrac{\Omega}{2}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\Omega}{2}-\dfrac{\Omega}{4}+k2\Omega=-\dfrac{3}{4}\Omega+k2\Omega\\3x=\dfrac{3}{2}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{3}{4}\Omega+k2\Omega\\x=\dfrac{1}{2}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)

c: \(sin2x+\sqrt{3}\cdot cosx=0\)

=>\(2\cdot sinx\cdot cosx+\sqrt{3}\cdot cosx=0\)

=>\(cosx\left(2\cdot sinx+\sqrt{3}\right)=0\)

=>\(\left[{}\begin{matrix}cosx=0\\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{2}+k\Omega\\x=-\dfrac{\Omega}{3}+k2\Omega\\x=\Omega+\dfrac{\Omega}{3}+k2\Omega=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)

d: \(sinx+sin2x+sin3x=0\)

=>\(sin2x+2\cdot sin\left(\dfrac{3x+x}{2}\right)\cdot cos\left(\dfrac{3x-x}{2}\right)=0\)

=>\(sin2x+2\cdot sin2x\cdot cosx=0\)

=>\(sin2x\left(2\cdot cosx+1\right)=0\)

=>\(\left[{}\begin{matrix}sin2x=0\\2\cdot cosx+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=k\Omega\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

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

Nguyễn Việt Lâm
3 tháng 8 2024 lúc 21:14

b.

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=-cosx\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=sin\left(x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=x-\dfrac{\pi}{2}+k2\pi\\2x+\dfrac{\pi}{4}=\dfrac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3\pi}{4}+k2\pi\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Nguyễn Việt Lâm
3 tháng 8 2024 lúc 21:15

c.

\(\Leftrightarrow2sinx.cosx+\sqrt{3}cosx=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)

Nguyễn Việt Lâm
3 tháng 8 2024 lúc 21:17

d.

\(\Leftrightarrow sinx+sin3x+sin2x=0\)

\(\Leftrightarrow2sin2x.cosx+sin2x=0\)

\(\Leftrightarrow sin2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

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

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


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