\(\Leftrightarrow sin5x+sinx-\left(1-2sin^2x\right)=0\)

\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

\(sin5x+sinx+2sin^2x=1\)

\(\Leftrightarrow\left(sin5x+sinx\right)-\left(1-2sin^2x\right)=0\)

\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)

\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\end{matrix}\right.\)

Vậy...

\(sin5x+sinx+sin3x=0\)

\(\Leftrightarrow2sin3x.cos2x+sin3x=0\)

\(\Leftrightarrow sin3x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

cho em hỏi chỗ 2sin3x.cos2x biến đổi sao vậy ạ? Từ công thức nào vậy ạ?

\(\Leftrightarrow sin5x=-sin3x\)

\(\Leftrightarrow sin5x=sin\left(-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=-3x+k2\pi\\5x=\pi+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=k2\pi\\2x=\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{4}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) (\(k\in Z\))

\(\Leftrightarrow cos2x-cos4x+\left(3\sqrt{2}-1\right)cos2x-3=0\)

\(\Leftrightarrow-cos4x+3\sqrt{2}cos2x-3=0\)

\(\Leftrightarrow-2cos^22x+3\sqrt{2}cos2x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\sqrt{2}\left(loại\right)\\cos2x=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow sin\left(3x+\dfrac{2\pi}{3}\right)=-sinx\)

\(\Leftrightarrow sin\left(3x+\dfrac{2\pi}{3}\right)=sin\left(-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2\pi}{3}=-x+k2\pi\\3x+\dfrac{2\pi}{3}=\pi+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-\dfrac{2\pi}{3}+k2\pi\\2x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\) (\(k\in Z\))

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{4}-x+k2\pi\\2x-\dfrac{\pi}{6}=\dfrac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{36}+\dfrac{k2\pi}{3}\\x=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\) (\(k\in Z\))

Hàm xác định trên R khi với mọi x ta có:

\(2sin3x+2cos3x-m>0\)

\(\Leftrightarrow sin3x+cos3x>\dfrac{m}{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(3x+\dfrac{\pi}{4}\right)>\dfrac{m}{2}\)

\(\Rightarrow\dfrac{m}{2\sqrt{2}}< \min\limits_Rsin\left(3x+\dfrac{\pi}{4}\right)=-1\)

\(\Rightarrow m< -2\sqrt{2}\)

Hàm xác định trên R khi với mọi x ta có:

\(sin^6x+cos^6x+m.sinx.cosx>0\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x+\dfrac{m}{2}sin2x>0\)

\(\Leftrightarrow3sin^22x-2m.sin2x-4< 0\)

Đặt \(sin2x=t\in\left[-1;1\right]\Rightarrow3t^2-2mt-4< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3.f\left(-1\right)< 0\\3.f\left(1\right)< 0\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2m-1< 0\\-2m-1< 0\end{matrix}\right.\)

\(\Rightarrow-\dfrac{1}{2}< m< \dfrac{1}{2}\)

\(\Leftrightarrow sin\left(3x+45^0\right)=sin\left(-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+45^0=-x+k360^0\\3x+45^0=180^0+x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=45^0+k360^0\\2x=135^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=22,5^0+k90^0\\x=67,5^0+k180^0\end{matrix}\right.\) (\(k\in Z\))

d, Hàm số xác định khi:

\(\left\{{}\begin{matrix}cos\left(x+\dfrac{\pi}{4}\right)\ne0\\sinx.cosx+cos2x-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\\\dfrac{1}{2}sin2x+cos2x\ne3\end{matrix}\right.\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)

e, Hàm số xác định khi:

\(\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)