Bài 2: Phương trình lượng giác cơ bản

\(1,cos\left(2x+\dfrac{\pi}{6}\right)=0.\\ \Leftrightarrow2x+\dfrac{\pi}{6}=k\pi.\\ \Leftrightarrow x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2},k\in Z.\)

\(2,sin\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=1.\\ \Leftrightarrow\dfrac{\pi}{4}-\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi.\\ \Leftrightarrow-\dfrac{x}{2}=\dfrac{\pi}{4}+k2\pi.\\ \Leftrightarrow x=\dfrac{-\pi}{2}-k\pi,k\in Z.\)

\(3,sin\left(3x-1\right)=\dfrac{1}{2}.\\ \Leftrightarrow sin\left(3x-1\right)=sin\dfrac{\pi}{6}.\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{\pi}{6}+k2\pi.\\3x-1=\pi-\dfrac{\pi}{6}+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+1+k2\pi.\\3x=\dfrac{5}{6}\pi+1+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{1}{3}+k\dfrac{2}{3}\pi.\\x=\dfrac{5}{18}\pi+\dfrac{1}{3}+k\dfrac{2}{3}\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(4,sin4x=sin\left(x+60^o\right).\\ \Leftrightarrow sin4x=sin\left(x+\dfrac{\pi}{3}\right).\Leftrightarrow\left[{}\begin{matrix}4x=x+\dfrac{\pi}{3}+k2\pi.\\4x=\pi-x-\dfrac{\pi}{3}+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{9}+k\dfrac{2}{3}\pi.\\x=\dfrac{2}{15}\pi+k\dfrac{2}{5}\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(5,cos\left(x-\dfrac{5\text{​​}\pi}{4}\right)=cos\left(2x+\text{​​}\dfrac{\pi}{4}\right).\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5\text{​​}\pi}{4}=2x+\text{​​}\dfrac{\pi}{4}+k2\pi.\\x-\dfrac{5\text{​​}\pi}{4}=-2x-\text{​​}\dfrac{\pi}{4}+k2\pi.\end{matrix}\right.\)

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

\(6,sin2x=cos\left(x+\text{​​}\dfrac{\text{​​}\pi}{3}\right).\\ sin2x=sin\left(\dfrac{\pi}{6}-x\right).\\ \Leftrightarrow\left[{}\begin{matrix}2x=\text{​​}\text{​​}\dfrac{\pi}{6}-x+k2\pi.\\2x=\pi-\dfrac{\pi}{6}+x+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+k\dfrac{2}{3}\pi.\\x=\dfrac{5}{6}\pi+k2\pi.\end{matrix}\right.\)\(\left(k\in Z\right).\)

\(\Leftrightarrow3x-\dfrac{\Pi}{4}=k2\Pi\)

hay \(x=\dfrac{1}{3}\left(k2\Pi+\dfrac{\Pi}{4}\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{\pi}{6}+k\pi\\x=\dfrac{1}{2}-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k3\pi\\x=-\dfrac{3\pi}{2}+k3\pi\end{matrix}\right.\)

Chọn C

\(2sin\left(x-\dfrac{\pi}{3}\right)-1=0\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

Do \(x\in\left[0;\pi\right]\Rightarrow x=\dfrac{\pi}{2}\)

Cả 4 đáp án của đề bài đều sai

\(\Leftrightarrow cos9x-cosx+sin10x=0\)

\(\Leftrightarrow-2sin5x.sin4x+2sin5x.cos5x=0\)

\(\Leftrightarrow sin5x\left(cos5x-sin4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\cos5x=sin4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\cos5x=cos\left(\dfrac{\pi}{2}-4x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=k\pi\\5x=\dfrac{\pi}{2}-4x+k2\pi\\5x=4x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{5}\\x=\dfrac{\pi}{18}+k2\pi.9\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

1.

\(2sin^2x+sin5x-sin3x=0\)

\(\Leftrightarrow2sin^2x+2cos4x.sinx=0\)

\(\Leftrightarrow2sinx\left(sinx+cos4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos4x=-sinx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos4x=cos\left(\dfrac{\pi}{2}+x\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

2.

\(sin3x+sinx-2sin^2x=0\)

\(\Leftrightarrow2sin2x.cosx-2sin^2x=0\)

\(\Leftrightarrow4sinx.cos^2x-2sin^2x=0\)

\(\Leftrightarrow4sinx\left(1-sin^2x\right)-2sin^2x=0\)

\(\Leftrightarrow2sinx\left(2-2sin^2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{-1+\sqrt{17}}{4}\\sinx=\dfrac{-1-\sqrt{17}}{4}< -1\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=arcsin\left(\dfrac{-1+\sqrt{17}}{4}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{-1+\sqrt{17}}{4}\right)+k2\pi\end{matrix}\right.\)

3.

\(cos5x-cosx+\sqrt{3}sin3x=0\)

\(\Leftrightarrow-2sin3x.sin2x+\sqrt{3}sin3x=0\)

\(\Leftrightarrow sin3x\left(-2sin2x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sin2x=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

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

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

1: \(\Leftrightarrow\sin^23x=\sin^2x\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin3x=\sin x\\\sin3x=\sin\left(-x\right)\end{matrix}\right.\)

TH1: sin 3x=sin x

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

5: \(\Leftrightarrow\Pi\cdot\sin\left(x-\dfrac{\Pi}{3}\right)=\dfrac{\Pi}{2}+k\Pi\)

\(\Leftrightarrow\sin\left(x-\dfrac{\Pi}{3}\right)=\dfrac{1}{2}+k\)

Để pt có nghiệm thì k+1/2>=-1 và k+1/2<=1

=>-3/2<=k<=1/2

Khi đó, pt sẽ có nghiệm là:

\(\left\{{}\begin{matrix}x-\dfrac{\Pi}{3}=arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\\x-\dfrac{\Pi}{3}=\Pi-arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi+\dfrac{\Pi}{3}\\x=\dfrac{4}{3}\Pi+arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow1-2x=k2\Pi\)

hay \(x=\dfrac{1-k2\Pi}{2}\)

1 - 2x = k2II

hoặc x = \(\dfrac{1-k2II}{2}\)