1) Giai phuong trinh:
a) Sin ( 4x + 10o) = Sin ( x - 20o )
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NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2021
\(\Leftrightarrow2cos^22x+3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)=2\)
\(\Leftrightarrow4cos^22x-3cos2x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=k2\pi\\2x=\pm arccos\left(-\dfrac{1}{4}\right)+k2\pi\\\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}arccos\left(-\dfrac{1}{4}\right)+k\pi\end{matrix}\right.\)
HP
20 tháng 9 2021
\(2cos^22x+3sin^2x=2\)
\(\Leftrightarrow-2\left(1-cos^22x\right)+3sin^2x=0\)
\(\Leftrightarrow-2sin^2x+3sin^2x=0\)
\(\Leftrightarrow sin^2x=0\)
\(\Leftrightarrow x=k\pi\)
HP
20 tháng 9 2021
Nguyễn Thái Sơn
\(\Leftrightarrow-2sin^22x+3sin^2x=0\)
\(\Leftrightarrow-2sin^22x+3sin^2x=0\)
\(\Leftrightarrow4sin^2x.cos^2x-3sin^2x=0\)
\(\Leftrightarrow sin^2x.\left(4cos^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\cos^2x=\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\pm\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)
...
HP
20 tháng 9 2021
\(3cos^2x-2sinx+2=0\)
\(\Leftrightarrow-3\left(1-cos^2x\right)-2sinx+5=0\)
\(\Leftrightarrow3sin^2x+2sinx-5=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(3sinx+5\right)=0\)
\(\Leftrightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 9 2021
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{7\pi}{24}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2019
a/ \(\Leftrightarrow2cosx.cos2x=cos2x\)
\(\Leftrightarrow2cosx.cos2x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
b/ \(\Leftrightarrow2sinx.sin2x=sinx\)
\(\Leftrightarrow2sinx.sin2x-sinx=0\)
\(\Leftrightarrow sinx\left(2sin2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\\sin2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=\frac{\pi}{6}+k2\pi\\2x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2019
c/ \(\Leftrightarrow sin3x-sinx+sin4x-sin2x=0\)
\(\Leftrightarrow2cos2x.sinx+2cos3x.sinx=0\)
\(\Leftrightarrow sinx\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow2sinx.2cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{5}+\frac{k4\pi}{5}\\x=\pi+k4\pi\end{matrix}\right.\)
d/ \(\Leftrightarrow sin3x-sinx-\left(sin4x-sin2x\right)=0\)
\(\Leftrightarrow2cos2x.sinx-2cos3x.sinx=0\)
\(\Leftrightarrow sinx\left(cos2x-cos3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=cos3x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=3x+k2\pi\\2x=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2020
\(\Leftrightarrow16sin^4x.cos^4x+cos^4x-1=0\)
\(\Leftrightarrow16sin^4x.cos^4x+\left(cos^2x+1\right)\left(cos^2x-1\right)=0\)
\(\Leftrightarrow16sin^4x.cos^4x-sin^2x\left(cos^2x+1\right)=0\)
\(\Leftrightarrow sin^2x\left(16sin^2x.cos^4x-cos^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\16sin^2x.cos^4x-cos^2x-1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow16cos^4x\left(1-cos^2x\right)-cos^2x-1=0\)
Đặt \(cos^2x=t\in\left[0;1\right]\)
\(\Rightarrow16t^2\left(1-t\right)-t-1=0\)
\(\Leftrightarrow-16t^3+16t^2-t-1=0\)
Nghiệm của pt bậc 3 này rất xấu cho nên chúng ta chỉ xác định được 1 nghiệm \(x=k\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+10^0=x-20^0+k360^0\\4x+10^0=200^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-30^0+k360^0\\5x=190^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-10^0+k120^0\\x=38^0+k72^0\end{matrix}\right.\) (\(k\in Z\))