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a.
\(sin4x+\sqrt{3}cos4x=-\sqrt{2}\)
\(\Leftrightarrow\frac{1}{2}sin4x+\frac{\sqrt{3}}{2}cos4x=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{3}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=-\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{3}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(2sin2x+2sin^2x=1\)
\(\Leftrightarrow2sin2x+1-cos2x=1\)
\(\Leftrightarrow2sin2x=cos2x\)
\(\Leftrightarrow tan2x=\frac{1}{2}\)
\(\Leftrightarrow2x=arctan\left(\frac{1}{2}\right)+k\pi\)
\(\Leftrightarrow...\)
c.
\(cos^2x-sin^2x-\sqrt{3}sin2x=1\)
\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\)
\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
d.
\(5sin2x-3\left(1+cos2x\right)=13\)
\(\Leftrightarrow5sin2x-3cos2x=16\)
Do \(5^2+\left(-3\right)^2< 16^2\) nên pt vô nghiệm
e.
\(\Leftrightarrow sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{2}\right)=cos\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow...\)
Từ phương trình ban đầu ta có : \(2\cos5x\sin x=\sqrt{3}\sin^2x+\sin x\cos x\)
\(\Leftrightarrow\begin{cases}\sin x=0\\2\cos5x=\sqrt{3}\sin x+\cos x\end{cases}\)
+) \(\sin x=0\Leftrightarrow x=k\pi\)
+)\(2\cos5x=\sqrt{3}\sin x+\cos x\Leftrightarrow\cos5x=\cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\begin{cases}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{18}+\frac{k\pi}{3}\end{cases}\)
\(\Leftrightarrow4sin4x.cos4x=\sqrt{2}\)
\(\Leftrightarrow2sin8x=\sqrt{2}\)
\(\Leftrightarrow sin8x=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}8x=\frac{\pi}{4}+k2\pi\\8x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{32}+\frac{k\pi}{4}\\x=\frac{3\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)