g)\(5sin2x-6cos^2x=13\)
\(\Leftrightarrow10.sinx.cosx-6cos^2x=13\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow13.sin^2x-10.sinx.cosx+19.cos^2x=0\)
\(\Leftrightarrow13\left(sinx-\dfrac{5}{13}cosx\right)^2+\dfrac{222}{13}cos^2x=0\) (vô nghiệm vì dấu bằng xảy ra khi sinx=cosx=0 \(\Rightarrow∄x\))
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h)\(cos^2x+2\sqrt{3}sinx.cosx+3sin^2x=1\)
\(\Leftrightarrow1+2sin^2x+2\sqrt{3}.sinx.cosx=1\)
\(\Leftrightarrow2sinx\left(sinx+\sqrt{3}cosx\right)=0\)
\(\Leftrightarrow2.sinx.2sin\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin\left(x+\dfrac{\pi}{3}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)(k nguyên)
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j) \(sinx+cosx=\sqrt{2}.sin4x\)
\(\Leftrightarrow sinx.\dfrac{1}{\sqrt{2}}+cosx.\dfrac{1}{\sqrt{2}}=sin4x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=4x+k2\pi\\x+\dfrac{\pi}{4}=\pi-4x+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}-\dfrac{k2\pi}{3}\\x=\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))
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