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<=> 2cos (3x/2)cos(x/2+pi/3)=0
<=>cos (3x/2)=0 hoặc cos (x/2+pi/3)=0
<=>3x/2=pi/2+kpi hoặc x/2+pi/3=pi/2+kpi (k thuộc z)
<=>x=pi/3+(2/3)kpi hoặc x=pi/3+2kpi (k thuộc z)
KL
\(\cos2x-\sin x+\cos x=0\Leftrightarrow\cos^2x-\sin^2x+\left(\cos x-\sin x\right)=0\)
\(\Leftrightarrow\left(\cos x-\sin x\right)\left(\cos x+\sin x+1\right)=0\)
\(\Leftrightarrow\begin{cases}\cos x-\sin x=0\\\cos x+\sin x+1=0\end{cases}\) \(\Leftrightarrow\begin{cases}\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)=0\\\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)=-1\end{cases}\)
\(\Leftrightarrow\begin{cases}x+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x-\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{cases}\) \(\Leftrightarrow\begin{cases}x=\frac{\pi}{4}+k\pi\\x=\pi+k2\pi\\x=-\frac{\pi}{2}+k2\pi\end{cases}\)
\(\Leftrightarrow2cos^2x-1+2cosx-\left(\dfrac{1}{2}-\dfrac{1}{2}cosx\right)=0\)
\(\Leftrightarrow2cos^2x+\dfrac{5}{2}cosx-\dfrac{3}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{-5+\sqrt{73}}{8}\\cosx=\dfrac{-5-\sqrt{73}}{8}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm arccos\left(\dfrac{-5+\sqrt{73}}{8}\right)+k2\pi\)
1: \(P=sin^22x=1-cos^22x\)
\(=1-\left(cos2x\right)^2\)
\(=1-\left(2cos^2x-1\right)^2\)
\(=1-\left(2\cdot\dfrac{9}{16}-1\right)^2\)
\(=1-\left(\dfrac{9}{8}-1\right)^2=1-\left(\dfrac{1}{8}\right)^2=\dfrac{63}{64}\)
2:
\(cos2x-sin\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(sin\left(x+\dfrac{\Omega}{3}\right)=cos2x=sin\left(\dfrac{\Omega}{2}-2x\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=\dfrac{\Omega}{2}-2x+k2\Omega\\x+\dfrac{\Omega}{3}=\Omega-\dfrac{\Omega}{2}+2x+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{\Omega}{6}+k2\Omega\\-x=\dfrac{1}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{18}+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{6}\Omega-k2\Omega\end{matrix}\right.\)
\(4\sin3x+\sin5x-2\sin x\cos2x=0\)
\(\Leftrightarrow\)\(4\sin3x+\sin5x-\sin3x+\sin x=0\)
\(\Leftrightarrow3\sin3x+\sin5x+\sin x=0\)
\(\Leftrightarrow3\sin3x+2\sin3x\cos2x=0\)
\(\Leftrightarrow\sin3x\left(3+2\cos2x\right)=0\)
Đáp số : \(x=k\dfrac{\pi}{3},k\in\mathbb{Z}\)
<=> cos2x=-sin(x-2pi/3)
<=>cos2x=sin(2pi/3-x)
<=>cos2x=cos(pi/2-2pi/3+x)
<=>cos2x=cos(x-pi/6)
=>2x=x-pi/6+k2pi và 2x=pi/6-x+k2pi (k thuộc z)
=>x=-pi/12+kpi và x=pi/18+(2/3)kpi