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a. vs m + 2
=>pttt : cos3x.cosx-sin2x+sin3xsinx+1=0
<=>\(\dfrac{1}{2}\left(cos2x+cos4x+cos2x-cos4x\right)-sin2x+1\)=0
<=>\(\dfrac{1}{2}\).2cos2x-sin2x+1=0
<=>cos2x-sin2x+1=0
<=>cos2x-sin2x-2sinxcosx+1=0
<=>cos2x+cos2x-sin2x=0
<=>2cos2x-2sinxcosx=0
<=>2cosx(cosx-sinx)=0
<=>\(\left[{}\begin{matrix}2cosx=0\\cosx-sinx=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{4+k\pi}\end{matrix}\right.\)(k thuộc Z)



tanx = 1-cos2x (ĐK x\(\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow\dfrac{sinx}{cosx}=2sin^2x\)
\(\Leftrightarrow sinx=2sin^2x\)
\(\Leftrightarrow sinx\left(2sinxcosx-1\right)\)=0
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

Ta có : \(\cos3x-\cos11x\)
\(=\cos9x+\cos3x\)
\(=\cos11x\)
\(=\cos\left(\pi-9x\right)\)



Lời giải:
Đặt $\sin x=a; \cos x=b(a,b\in [-1;1])$ thì ta có:
\(\left\{\begin{matrix} 3a+b+2=0\\ a^2+b^2=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} b=-(3a+2)\\ a^2+b^2=1\end{matrix}\right.\)
\(\Rightarrow a^2+(3a+2)^2=1\)
\(\Leftrightarrow 10a^2+12a+3=0\Rightarrow a=\frac{-6\pm \sqrt{6}}{10}\)
Với $a=\frac{-6\pm \sqrt{6}}{10}$ thì \(x=2k\pi +\arcsin \frac{-6+\sqrt{6}}{10}\) hoặc \(x=(2k+1)\pi -\arcsin \frac{-6+\sqrt{6}}{10}\)Với $a=\frac{-6-\sqrt{6}}{10}$ thì \(x=2k\pi +\arcsin \frac{-6-\sqrt{6}}{10}\) hoặc \(x=(2k+1)\pi -\arcsin \frac{-6-\sqrt{6}}{10}\)
Với $k$ nguyên nào đó.
g, \(1+2sinx=2cosx\)
\(\Leftrightarrow sinx-cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x-\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
h, \(4cosx-3sinx=3\)
\(\Leftrightarrow5\left(\dfrac{4}{5}cosx-\dfrac{3}{5}sinx\right)=3\)
\(\Leftrightarrow cos\left(x+arccos\dfrac{4}{5}\right)=\dfrac{3}{5}\)
\(\Leftrightarrow x+arccos\dfrac{4}{5}=\pm arccos\dfrac{3}{5}+k2\pi\)
\(\Leftrightarrow x=-arccos\dfrac{4}{5}\pm arccos\dfrac{3}{5}+k2\pi\)