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\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt[]{3}sin2x-cos2x\)
\(\Leftrightarrow4.\left(-\dfrac{1}{2}\right)\left[sin\left(x+\dfrac{\pi}{3}+x-\dfrac{\pi}{6}\right)+sin\left(x+\dfrac{\pi}{3}-x+\dfrac{\pi}{6}\right)\right]=m^2+2.\left[\dfrac{\sqrt[]{3}}{2}.sin2x-\dfrac{1}{2}.cos2x\right]\)
\(\Leftrightarrow2\left[sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(2x-\dfrac{\pi}{6}\right)\right]=m^2+2\)
\(\Leftrightarrow2.2sin2x.cos\dfrac{\pi}{6}=m^2+2\)
\(\Leftrightarrow2.2sin2x.\dfrac{\sqrt[]{3}}{2}=m^2+2\)
\(\Leftrightarrow2\sqrt[]{3}sin2x.=m^2+2\)
\(\Leftrightarrow sin2x.=\dfrac{m^2+2}{2\sqrt[]{3}}\)
Phương trình có nghiệm khi và chỉ khi
\(\left|\dfrac{m^2+2}{2\sqrt[]{3}}\right|\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m^2+2}{2\sqrt[]{3}}\ge-1\\\dfrac{m^2+2}{2\sqrt[]{3}}\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2\ge-2\left(1+\sqrt[]{3}\right)\left(luôn.đúng\right)\\m^2\le2\left(1-\sqrt[]{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt[]{2\left(1-\sqrt[]{3}\right)}\le m\le\sqrt[]{2\left(1-\sqrt[]{3}\right)}\)
PT \(\Leftrightarrow-2\left(1-2.sin^2\dfrac{x}{2}\right)-\sqrt{3}.cos2x=-1+2\left(cosx.cos\dfrac{3\pi}{4}-sinx.sin\dfrac{3\pi}{4}\right)^2\)
\(\Leftrightarrow-2.cosx-\sqrt{3}.cos2x=-1+2\left(cosx.-\dfrac{\sqrt{2}}{2}-sinx.\dfrac{\sqrt{2}}{2}\right)^2\)
\(\Leftrightarrow-2cosx-\sqrt{3}.cos2x=-1+\left(sinx+cosx\right)^2\)
\(\Leftrightarrow-2cosx=2sinx.cosx+\sqrt{3}cos2x\)
\(\Leftrightarrow-2cosx=sin2x+\sqrt{3}cos2x\)
\(\Leftrightarrow cos\left(\pi-x\right)=\dfrac{1}{2}.sin2x+\dfrac{\sqrt{3}}{2}.cos2x\)
\(\Leftrightarrow cos\left(\pi-x\right)=sin\left(2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(\pi-x\right)=cos\left(\dfrac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi-x=\dfrac{\pi}{6}-2x+k2\pi\\\pi-x=-\dfrac{\pi}{6}+2x+k2\pi\end{matrix}\right.\) ( k nguyên )
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{7\pi}{18}-\dfrac{k2\pi}{3}\end{matrix}\right.\) ( k nguyên )
Vậy...
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
\(\Leftrightarrow2\sqrt{2}cos2x+sin2x\left(cosx.cos\left(\frac{3\pi}{4}\right)-sinx.sin\left(\frac{3\pi}{4}\right)\right)-2\sqrt{2}\left(sinx+cosx\right)=0\)
\(\Leftrightarrow2\left(cos^2x-sin^2x\right)-sinx.cosx\left(cosx+sinx\right)-2\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx-2sinx-sinx.cosx-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\Rightarrow...\\2\left(cosx-sinx\right)-sinx.cosx-2=0\left(1\right)\end{matrix}\right.\)
Xét (1)
Đặt \(cosx-sinx=t\Rightarrow sinx.cosx=\frac{1-t^2}{2}\) (với \(\left|t\right|\le\sqrt{2}\))
\(\Rightarrow2t-\frac{1-t^2}{2}-2=0\Leftrightarrow t^2+4t-5=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-5\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow cosx-sinx=1\Leftrightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\Leftrightarrow...\)