1.
\(2tan^2x-5tanx+3=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(2tanx-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{3}{2}\right)+k\pi\end{matrix}\right.\)
2.
\(25sin^2x+15cosx+10=0\)
\(\Leftrightarrow25\left(1-cos^2x\right)+15cosx+10=0\)
\(\Leftrightarrow-25cos^2x+15cosx+35=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\dfrac{3+\sqrt{149}}{10}>1\left(loại\right)\\cosx=\dfrac{3-\sqrt{149}}{10}\end{matrix}\right.\)
\(\Rightarrow x=\pm arccos\left(\dfrac{3-\sqrt{149}}{10}\right)+k2\pi\)
3.
\(2sinx+cosx=1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{5}}cosx+\dfrac{2}{\sqrt{5}}sinx=\dfrac{1}{\sqrt{5}}\)
Đặt \(\dfrac{1}{\sqrt{5}}=cosa\Rightarrow\dfrac{2}{\sqrt{5}}=sina\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)
\(\Rightarrow cosx.cosa+sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x-a\right)=cosa\)
\(\Rightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=-a+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=k2\pi\end{matrix}\right.\)
4.
\(\sqrt{3}sinx+cosx=\sqrt{2}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{6}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
5.
\(2sinx+2cosx-\sqrt{2}=0\)
\(\Leftrightarrow sinx+cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
