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+ k2π ⇔ x = .png)
+ kπ, (k ∈ Z).
ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)
\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)
\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)
\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)
\(\Leftrightarrow2cos^22x-2cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)
Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)
a) <=> 4sinxcosx -(2cos2x-1)=7sinx+2cosx-4
<=> 2cos2x+(2-4sinx)cosx+7sinx-5=0
- sinx=1 => 2cos2x-2cosx+2=0
pt trên vn
b) <=> 2sinxcosx-1+2sin2x+3sinx-cosx-1=0
<=> cos(2sinx-1)+2sin2x+3sinx-2=0
<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0
<=> (2sinx-1)(cosx+sinx+2)=0
<=> sinx=1/2 hoặc cosx+sinx=-2(vn)
<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)
\(sin2x-cos2x+3sinx-cosx-1=0\)
\(\Leftrightarrow2sinxcosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow2sinxcosx-1+2sin^2x+3sinx-cosx-1=0\)
\(\Leftrightarrow2sin^2x+3sinx-2+cosx\left(2sinx-1\right)=0\)
\(\Leftrightarrow2\left(sinx-\dfrac{1}{2}\right)\left(sinx+2\right)+cosx\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+2\right)+cosx\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+2+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx+cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=sin\dfrac{\pi}{6}\\\sqrt[]{2}\left(sinx.\dfrac{1}{\sqrt[]{2}}+cosx.\dfrac{1}{\sqrt[]{2}}\right)=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\\sqrt[]{2}sin\left(x+\dfrac{\pi}{4}\right)=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt[]{2}\left(vô.lý\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
a: ĐKXĐ: sin 2x<>1
=>2x<>pi/2+k2pi
=>x<>pi/4+kpi
\(\dfrac{cos2x}{sin2x-1}=0\)
=>cos2x=0
=>2x=pi/2+kpi
=>x=pi/4+kpi/2
Kết hợp ĐKXĐ, ta được:
x=3/4pi+k2pi hoặc x=7/4pi+k2pi
b: cos(sinx)=1
=>sin x=kpi
=>sin x=0
=>x=kpi
c: \(2\cdot sin^2x-1+cos3x=0\)
=>cos3x+cos2x=0
=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)
=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)
=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi
=>x=-pi+k2pi hoặc x=pi/5+k2pi/5
e: cos3x=-cos7x
=>cos3x=cos(pi-7x)
=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi
=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2
1.
ĐK: \(x\ne\dfrac{\pi}{4}+k\pi\)
\(\dfrac{cos2x}{1-sin2x}=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Đối chiếu điều kiên ta được \(x=-\dfrac{\pi}{4}+k\pi\)
2.
ĐK: \(x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3};x\ne\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
\(tan3x=tan4x\)
\(\Leftrightarrow3x=4x+k\pi\)
\(\Leftrightarrow x=k\pi\)