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b.
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=-\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=-\pi+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\dfrac{3}{5}sinx-\dfrac{4}{5}cosx=1\)
Đặt \(\dfrac{3}{5}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{4}{5}=sina\)
Pt trở thành:
\(sinx.cosa-cosx.sina=1\)
\(\Leftrightarrow sin\left(x-a\right)=1\)
\(\Leftrightarrow x-a=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=a+\dfrac{\pi}{2}+k2\pi\)
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33.
\(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x=cosx\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cosx\)
So sánh nó với \(cos\left(2x-a\right)=cosx\)
\(\Rightarrow a=\dfrac{\pi}{3}\)
34.
ĐKXĐ:
\(sinx-cosx\ne0\)
\(\Leftrightarrow tanx\ne1\)
\(\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)
35.
\(y=2\left(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx\right)-2=2sin\left(x-\dfrac{\pi}{6}\right)-2\)
Do \(-1\le sin\left(x-\dfrac{\pi}{6}\right)\le1\Rightarrow-4\le y\le0\)
Tập giá trị: \(\left[-4;0\right]\)
36.
\(y=cos2x\) tuần hoàn chu kì \(\dfrac{2\pi}{\left|2\right|}=\pi\)
\(y=sinx\) tuàn hoàn chu kì \(\dfrac{2\pi}{\left|1\right|}=2\pi\)
\(y=tan2x\) tuần hoàn chu kì \(\dfrac{\pi}{\left|2\right|}=\dfrac{\pi}{2}\)
\(y=cot4x\) tuần hoàn chu kì \(\dfrac{\pi}{\left|4\right|}=\dfrac{\pi}{4}\)
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1. ĐKXĐ:
a. \(sin\left(x+\dfrac{\pi}{3}\right)\ne0\Rightarrow x+\dfrac{\pi}{3}\ne k\pi\Rightarrow x\ne-\dfrac{\pi}{3}+k\pi\)
b. \(cos\left(x-\dfrac{\pi}{4}\right)\ne0\Rightarrow x-\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{3\pi}{4}+k\pi\)
2.
a.
\(-1\le sinx\le1\Rightarrow-5\le-5sinx\le5\Rightarrow-3\le y\le8\)
Tập giá trị: \(\left[-3;8\right]\)
b.
\(-1\le cos2x\le1\Rightarrow-4\le4cos2x\le4\Rightarrow3\le y\le11\)
Tập giá trị: \(\left[3;11\right]\)
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\(2cos\left(x+\dfrac{\pi}{6}\right)-1=0\Rightarrow cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+n2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=-\dfrac{\pi}{2}+n2\pi\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-\dfrac{\pi}{2}\le\dfrac{\pi}{6}+k2\pi< 2\pi\\-\dfrac{\pi}{2}\le-\dfrac{\pi}{2}+n2\pi< 2\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{3}\le k< \dfrac{11}{12}\\0\le n< \dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=\left\{0\right\}\\n=\left\{0;1\right\}\end{matrix}\right.\) \(\Rightarrow x=\left\{\dfrac{\pi}{6};-\dfrac{\pi}{2};\dfrac{3\pi}{2}\right\}\Rightarrow\dfrac{\pi}{6}-\dfrac{\pi}{2}+\dfrac{3\pi}{2}=\dfrac{7\pi}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3.
\(4sinx+cosx+2cos\left(x+\dfrac{\pi}{3}\right)=2\)
\(\Leftrightarrow4sinx+cosx+cosx-\sqrt{3}sinx=2\)
\(\Leftrightarrow\left(4-\sqrt{3}\right)sinx+2cosx=2\)
\(\Leftrightarrow\sqrt{23-4\sqrt{3}}\left(\dfrac{4-\sqrt{3}}{\sqrt{23-4\sqrt{3}}}sinx+\dfrac{2}{\sqrt{23-4\sqrt{3}}}cosx\right)=2\)
\(\Leftrightarrow cos\left(x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}\right)=\dfrac{2}{\sqrt{23-4\sqrt{3}}}\)
\(\Leftrightarrow x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}=\pm arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)
4.
\(sinx+2cos\left(x+\dfrac{\pi}{3}\right)+4sin\left(x+\dfrac{\pi}{6}\right)+cosx=4\)
\(\Leftrightarrow sinx+cosx-\sqrt{3}sinx+2\sqrt{3}sinx+2cosx+cosx=4\)
\(\Leftrightarrow\left(1+\sqrt{3}\right)sinx+4cosx=4\)
\(\Leftrightarrow\sqrt{20+2\sqrt{3}}\left(\dfrac{1+\sqrt{3}}{\sqrt{20+2\sqrt{3}}}sinx+\dfrac{4}{\sqrt{20+2\sqrt{3}}}cosx\right)=4\)
\(\Leftrightarrow cos\left(x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}\right)=\dfrac{4}{\sqrt{20+2\sqrt{3}}}\)
\(\Leftrightarrow x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}=\pm arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)