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1/ \(sinx=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
b/ \(cos=-\frac{\sqrt{2}}{2}=cos\left(\frac{3\pi}{4}\right)\)
\(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
c/ \(tanx=\sqrt{3}=tan\left(\frac{\pi}{3}\right)\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
d/ \(cotx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)
2/
a/ \(sin^2x+sinx-2=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-2\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
b/ \(cot^2x-2cotx-3=0\)
\(\Leftrightarrow\left(cotx+1\right)\left(cotx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot3+k\pi\end{matrix}\right.\)
3/ \(\Leftrightarrow1-cos2x+1-cos4x+1-cos6x=3\)
\(\Leftrightarrow cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2coss4x.cos2x+cos4x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\frac{2\pi}{3}+k2\pi\\2x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow\left(sinx-1\right)\left(sinx-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=4\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
b/
\(\Leftrightarrow\left(cos2x-1\right)\left(2cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sin3x-\frac{3}{4}\right)^2+\frac{7}{16}=0\)
Vế trái luôn dương nên pt vô nghiệm
e.
\(3\left(1-sin^2x\right)-5sinx-1=0\)
\(\Leftrightarrow-3sin^2x-5sinx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{3}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
f.
\(2\left(2cos^2x-1\right)-cosx+7=0\)
\(\Leftrightarrow4cos^2x-cosx+5=0\)
Phương trình vô nghiệm
g.
\(\Leftrightarrow\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)=2\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=\sqrt{2}>1\)
Phương trình vô nghiệm
h.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
a, ta có 2x + π/3 = 3π/4 +k2π hoặc 2x + π/3 = -3π/4 + k2π
=> x= 5π/24 + kπ hoặc x= -13π/24 +kπ
b, đề sai phải ko
c, cos22x - sin22x - 2sinx -1=0
<=> -2sin22x -2sin2x =0
<=> sin2x=0 hoặc sin2x=-1
<=> x=kπ hoặc x= π/2 + kπ ; x=-π/4 +kπ hoặc x=5π/8 + kπ
d, cos5xcosπ/4 - sin5xsinπ/4 = -1/2
cos( 5x + π/4 ) = -1/2
<=> x=π/12 +k2π/5 hoặc x= -11π/60 + k2π/5
f,4x+π/3=3π/10 -x +k2π hoặc 4x+π/3 = x - 3π/10 +k2π
<=> x =-π/150 + k2π/5 hoặc x = π/90 +k2π/3
ĐKXĐ: ...
a/ \(\frac{sin2x}{cos2x}+\frac{cosx}{sinx}=8cos^2x\)
\(\Leftrightarrow sin2x.sinx+cos2x.cosx=8cos^2x.sinx.cos2x\)
\(\Leftrightarrow cosx=4sin2x.cos2x.cosx\)
\(\Leftrightarrow cosx=2sin4x.cosx\)
\(\Leftrightarrow cosx\left(2sin4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin4x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
b/ \(\frac{cosx}{sinx}-\frac{sinx}{cosx}+4sin2x=\frac{1}{sinx.cosx}\)
\(\Leftrightarrow cos^2x-sin^2x+4sin2x.sinx.cosx=1\)
\(\Leftrightarrow cos2x+2sin^22x=1\)
\(\Leftrightarrow cos2x+2\left(1-cos^22x\right)=1\)
\(\Leftrightarrow-2cos^22x+cos2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
1c/
\(5sinx-2=3\left(1-sinx\right)\frac{sin^2x}{1-sin^2x}\)
\(\Leftrightarrow5sinx-2=\frac{3sin^2x}{1+sinx}\)
\(\Leftrightarrow\left(5sinx-2\right)\left(1+sinx\right)=3sin^2x\)
\(\Leftrightarrow5sin^2x+3sinx-2=3sin^2x\)
\(\Leftrightarrow2sin^2x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=...\)
Bài 2:
a/ \(\Leftrightarrow\frac{\left(m+1\right)\left(1-cos2x\right)}{2}-sin2x+cos2x=0\)
\(\Leftrightarrow2sin2x+\left(m-1\right)cos2x=m+1\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(4+\left(m-1\right)^2\ge\left(m+1\right)^2\)
\(\Leftrightarrow4m\le4\Rightarrow m\le1\)
c/
\(\Leftrightarrow1-cos^2\frac{x}{2}-2cos\frac{x}{2}+2=0\)
\(\Leftrightarrow cos^2\frac{x}{2}+2cos\frac{x}{2}-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=1\\cos\frac{x}{2}=-3< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{2}=k2\pi\)
\(\Leftrightarrow x=k4\pi\)
d/ ĐKXĐ: ...
\(\Leftrightarrow tanx-\frac{2}{tanx}+1=0\)
\(\Leftrightarrow tan^2x+tanx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow\left(cosx-1\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
b \(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\pm\frac{3\pi}{8}+k\pi\end{matrix}\right.\)