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d: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
a: =>sin 4x=cos(x+pi/6)
=>sin 4x=sin(pi/2-x-pi/6)
=>sin 4x=sin(pi/3-x)
=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi
=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3
b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi
=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi
c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
Pt \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=2x+\dfrac{\pi}{3}+k2\pi\\\dfrac{\pi}{3}-x=-2x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-k2\pi}{3}\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
1: cos(2x+pi/6)=cos(pi/3-3x)
=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi
=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi
=>x=pi/30+k2pi/5 hoặc x=pi-k2pi
2: sin(2x+pi/6)=sin(pi/3-3x)
=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi
=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi
=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi
1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-3x+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)
\(cos\left(x-\dfrac{\pi}{3}\right)=sin\left(2x+\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{\pi}{3}-x+l2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{9}+l\dfrac{2\pi}{3}\end{matrix}\right.\)
Chỉ II đúng
a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{3}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k\pi\\2x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+x+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=cos\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+x+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{7\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)
c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)
=>\(cos\left(x-\dfrac{pi}{6}\right)=sin\left(-2x-\dfrac{pi}{3}\right)\)
=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(\dfrac{pi}{2}-x+\dfrac{pi}{6}\right)\)
=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(-x+\dfrac{2}{3}pi\right)\)
=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)
a: cos3x=8
mà -1<=cos3x<=1
nên \(x\in\varnothing\)
b; \(-2\cdot cosx+\sqrt{3}=0\)
=>\(-2\cdot cosx=-\sqrt{3}\)
=>\(cosx=\dfrac{\sqrt{3}}{2}\)
=>x=pi/6+k2pi hoặc x=-pi/6+k2pi
c: cos(3x-pi/6)=0
=>3x-pi/6=pi/2+k2pi
=>3x=2/3pi+k2pi
=>x=2/9pi+k2pi/3
d: cos(x+2/3pi)=cos(pi/5)
=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi
=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi
e: cos^2(3x)=4
=>cos3x=2(loại) hoặc cos3x=-2(loại)
<=> 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
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}cos4x+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}\left(1-2sin^22x\right)+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow...\)
\(cos\left(2x+\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow2cos\dfrac{3x}{2}.cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{3x}{2}=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)