Question

Giải PT

a1) \(3.\cos4x-2^{ }\cos^23x=1\)

a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)

a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)

a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)

Answer 1

a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)

\(\Leftrightarrow3cos4x-cos6x-2=0\)

Đặt \(t=2x\)

Pttt:\(3cos2t-cos3t-2=0\)

\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)

\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))

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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) (\(k\in Z\))

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a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)

Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)

\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)

\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

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Answer 2

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 \(6^2+2^2< 7^2\))

\(\Rightarrow sinx=1\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)

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