3: \(cos2x+3\sqrt{3}\cdot sin2x+4\cdot sin^2x=-5\)
=>\(cos2x+\sqrt{3}\cdot sin2x+2\sqrt{3}\cdot sin2x+4\cdot\dfrac{1-cos2x}{2}=-5\)
=>\(2\left(\dfrac{1}{2}\cdot cos2x+\dfrac{\sqrt{3}}{2}\cdot sin2x\right)+2\sqrt{3}\cdot sin2x+2-2cos2x=-5\)
=>\(cos2x+\sqrt{3}\cdot sin2x+2\sqrt{3}\cdot sin2x-2cos2x=-3\)
=>\(3\sqrt{3}\cdot sin2x-cos2x=-3\)
=>\(\dfrac{3\sqrt{3}}{\sqrt{28}}\cdot sin2x-\dfrac{1}{\sqrt{28}}\cdot cos2x=-\dfrac{3}{\sqrt{28}}\)
=>\(sin\left(2x-\alpha\right)=-\dfrac{3}{\sqrt{28}}\)(Với \(cos\alpha=\dfrac{3\sqrt{3}}{28};sin\alpha=\dfrac{1}{\sqrt{28}}\)
=>\(\left[{}\begin{matrix}2x-\alpha=arcsin\left(-\dfrac{3}{\sqrt{28}}\right)+k2\Omega\\2x-\alpha=\Omega-arcsin\left(-\dfrac{3}{\sqrt{28}}\right)+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\left[\alpha+arcsin\left(-\dfrac{3}{\sqrt{28}}\right)+k2\Omega\right]\\x=\dfrac{1}{2}\left(\Omega+\alpha-arcsin\left(-\dfrac{3}{\sqrt{28}}\right)+k2\Omega\right)\end{matrix}\right.\)
5: \(cos^2x-3\cdot sinx\cdot cosx+2\cdot sin^2x=0\)
=>\(cos^2x-sinx\cdot cosx-2\cdot sinx\cdot cosx+2\cdot sin^2x=0\)
=>\(cosx\cdot\left(cosx-sinx\right)-2\cdot sinx\left(cosx-sinx\right)=0\)
=>\(\left(cosx-sinx\right)\left(cosx-2\cdot sinx\right)=0\)
=>\(\left[{}\begin{matrix}cosx-sinx=0\\cosx-2\cdot sinx=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{2}}{2}\cdot sinx-\dfrac{\sqrt{2}}{2}\cdot cosx=0\\2\cdot sinx-cosx=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}sin\left(x-\dfrac{\Omega}{4}\right)=0\\sinx\cdot\dfrac{2}{\sqrt{5}}-cosx\cdot\dfrac{1}{\sqrt{5}}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x-\dfrac{\Omega}{4}=k\Omega\\sin\left(x-\alpha\right)=0\left(cos\alpha=\dfrac{2}{\sqrt{5}}\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k\Omega+\dfrac{\Omega}{4}\\x=k\Omega+\alpha\end{matrix}\right.\)
mà \(x\in\left(-2\Omega;2\Omega\right)\)
nên \(x\in\left\{-\dfrac{7}{4}\Omega;-\dfrac{3}{4}\Omega;\dfrac{\Omega}{4};\dfrac{5}{4}\Omega;\alpha;\alpha+\Omega;\alpha-\Omega\right\}\)
