giải pt: cosx = 1/2
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1.
ĐK: \(x\ne\dfrac{k\pi}{2}\)
\(cotx-tanx=sinx+cosx\)
\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)
\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=sinx+cosx\)
\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx=sinx.cosx\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)
\(\Leftrightarrow t^2+2t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)
\(\Leftrightarrow-\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-1+\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}-1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Để pt có nghiệm \(\Rightarrow cosx\ge0\)
Khi đó \(\Rightarrow\left(1+sinx\right)^4=cos^2x\)
\(\Rightarrow\left(1+sinx\right)^4=1-sin^2x\)
\(\Rightarrow\left(1+sinx\right)^4=\left(1+sinx\right)\left(1-sinx\right)\)
\(\Rightarrow\left[{}\begin{matrix}sinx+1=0\\\left(1+sinx\right)^3=1-sinx\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx\left(sin^2x+3sinx+4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\cosx=-1\left(loại\right)\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b/ ĐKXĐ: ...
\(\Leftrightarrow tan^2x+1-\frac{4}{cosx}+4=0\)
\(\Leftrightarrow\frac{1}{cos^2x}-\frac{4}{cosx}+4=0\)
\(\Leftrightarrow\left(\frac{1}{cosx}-2\right)^2=0\)
\(\Leftrightarrow\frac{1}{cosx}=2\)
\(\Rightarrow cosx=\frac{1}{2}\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
a/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3}\frac{sinx}{cosx}+1=\frac{1}{cos^2x}\)
\(\Leftrightarrow\sqrt{3}tanx+1=1+tan^2x\)
\(\Leftrightarrow tanx\left(tanx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
\(cosx=\dfrac{1}{2}\)
\(\Leftrightarrow cosx=cos\dfrac{\pi}{3}\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)