giải pt \(cos2\left(x+\frac{\pi}{6}\right)+4cos\left(\frac{\pi}{5}-x\right)=\frac{5}{2}\)
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DP
29 tháng 5 2017
\(cos2\left(x+\frac{\pi}{6}\right)+4cos\left(\frac{\pi}{3}-x\right)=\frac{5}{2}\)
\(4sin\left(x+\frac{\pi}{6}\right)+\left(x+\frac{\pi}{6}\right)cos2=\frac{5}{2}\)
\(\frac{1}{6}\left(24sin\right)\left(x+\frac{\pi}{6}\right)+6x\left(cos2\right)=\frac{5}{2}\)
\(2\sqrt{3}sin\left(x\right)+x\)\(cos\left(2\right)+2cos\left(x\right)+\frac{1}{6}\pi\)\(cos\left(2\right)=\frac{5}{2}\)
\(\left(2\sqrt[6]{-1}-2\left(-1^{\frac{5}{6}}\right)\right)sin\left(x\right)+x\left(cos2\right)+\left(2\sqrt[3]{-1-2\left(-1^{\frac{2}{3}}\right)}\right)cos\left(x\right)=\frac{5}{2}-\frac{1}{6}\pi\)\(cos\left(2\right)\)
\(24sin\left(x+\frac{\pi}{6}\right)+\left(6x+\pi\right)cos\left(2\right)=15\)
\(4sin\left(x+\frac{\pi}{6}\right)+x\)\(cos\left(2\right)+\frac{1}{6}\pi\)\(cos\left(2\right)=\frac{5}{2}\)
\(\Rightarrow x=\left\{-15,1252;-13,976;-6,8388;-3,93832\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
\(cos2\left(x+\frac{\pi}{3}\right)=2cos^2\left(x+\frac{\pi}{3}\right)-1=2cos^2\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]-1\)
\(=2sin^2\left(\frac{\pi}{6}-x\right)-1=2\left(1-cos^2\left(\frac{\pi}{6}-x\right)\right)-1=1-2cos^2\left(\frac{\pi}{6}-x\right)=1-2t^2\)
Vậy pt trở thành: \(1-2t^2+4t=\frac{5}{2}\Leftrightarrow2t^2-4t+\frac{3}{2}=0\)
AT
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 10 2020
\(cos^2\left(\frac{\pi}{3}+x\right)+4sin\left(x+\frac{\pi}{3}\right)-4=0\)
\(\Leftrightarrow1-sin^2\left(x+\frac{\pi}{3}\right)+4sin\left(x+\frac{\pi}{3}\right)-4=0\)
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)-4sin\left(x+\frac{\pi}{3}\right)+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=1\\sin\left(x+\frac{\pi}{3}\right)=3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
d/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2020
c/
\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2020
a/
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3}\)
b/
\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} = - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi + k2\pi }\\{3x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)
\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)
b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} = - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi + k2\pi }\\{x = - \pi + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
\(cos^2\left(x+\frac{\pi}{6}\right)\) hả
uk đúng rồi ...mk viết nhầm ..phải là cos^2 nha