(2 cos x +\(\sqrt{2}\))(cos x-2)=0
(tan x-\(\sqrt{3}\))(1- tan x)=0
(cot \(\frac{x}{3}\)-1)(cot \(\frac{x}{2}\)+1)=0
Giải mấy phương trình này giúp mik với mọi người ơi
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)
b) \(2\cos x = - \sqrt 2 \;\; \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x = - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)
c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)
\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)
d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
HP
25 tháng 6 2021
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
25 tháng 6 2021
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)
Vậy...
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
Vậy...
c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
Vậy...
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow cos2x=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
Vậy...
e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))
\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
Vậy...
11 tháng 7 2021
a) \(\left|sinx-cosx\right|+\left|sinx+cosx\right|=2\)
\(\Leftrightarrow\left(sinx-cosx\right)^2+2\left|sinx-cosx\right|\left|sinx+cosx\right|+\left(cosx+sinx\right)^2=4\)
\(\Leftrightarrow2\left(sin^2x+cos^2x\right)+2\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|=4\)
\(\Leftrightarrow\left|sin^2x-cos^2x\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=1\\sin^2x-cos^2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=sin^2x+cos^2x\\sin^2x-cos^2x=-\left(sin^2x+cos^2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=0\\sin^2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=0\end{matrix}\right.\)\(\Rightarrow cosx.sinx=0\Rightarrow sin2x=0\)
\(\Rightarrow x=\dfrac{k\pi}{2},k\in Z\)
Vậy...
b) ĐK:\(x\ne\dfrac{k\pi}{2};k\in Z\)
Pt \(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cosx}{sinx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{cosx.sinx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\dfrac{\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)}{sinx.cosx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\left(1\right)\\\dfrac{sinx-\sqrt{3}cosx}{sinx.cosx}=4\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\Leftrightarrow tanx=-\sqrt{3}\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi,k\in Z\)
Từ (2)\(\Leftrightarrow sinx-\sqrt{3}cosx=4sinx.cosx\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=2sinx.cosx\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin2x\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)
c) ĐK: \(x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)
Pt \(\Leftrightarrow\left(\sqrt{2}sinx-1\right)^2+\left(\sqrt{3}tan2x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}sinx-1=0\\\sqrt{3}tan2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sinx=\dfrac{1}{\sqrt{2}}\\tan2x=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy pt vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 11 2019
a/ Thiếu đề, sau dấu "-" hình như còn gì đó
b/ \(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=sin\left(\frac{\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/ \(\Rightarrow sin2x=-sinx\Leftrightarrow sin2x=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=-x+k2\pi\\2x=\pi+x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)
d/ \(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1\)
\(\Leftrightarrow sinx.cosx=0\Leftrightarrow sin2x=0\)
\(\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)
e/ f/ Thiếu đề
g/ \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=-cos2x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=cos\left(\pi-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=2x+k2\pi\\3x=-2x+k2\pi\\3x=\pi-2x+k2\pi\\3x=2x-\pi+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=-\pi+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 10 2020
a. ĐKXĐ: ...
\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 10 2020
b.
\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)
\(\Leftrightarrow4cos^32x-2cos2x-1=0\)
Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề
c. ĐKXĐ: ...
\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=...\\\frac{cosx+sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx+sinx=sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)
TG
2
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
b.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\sqrt{2}\left(sinx+cosx\right)=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\)
\(\Leftrightarrow\sqrt{2}\left(sinx+cosx\right)=\dfrac{1}{sinx.cosx}\)
Đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(sinx.cosx=\dfrac{t^2-1}{2}\)
Pt trở thành:
\(\sqrt{2}t=\dfrac{2}{t^2-1}\Rightarrow t^3-t-\sqrt{2}=0\)
\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+\sqrt{2}t+1\right)=0\)
\(\Leftrightarrow t=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
a.
\(\Leftrightarrow sin^22x+cos^22x+\sqrt{3}sin4x+1+cos4x=0\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-2\)
\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{\sqrt{3}}{2}sin4x=-1\)
\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{3}\right)=-1\)
\(\Leftrightarrow4x-\dfrac{\pi}{3}=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\)
Lời giải:
a.
$(2\cos x+\sqrt{2})(\cos x-2)=0$
\(\Rightarrow \left[\begin{matrix} 2\cos x+\sqrt{2}=0\\ \cos x-2=0\end{matrix}\right.\)
Nếu $2\cos x+\sqrt{2}=0\Rightarrow \cos x=\frac{-\sqrt{2}}{2}\Rightarrow x=\pm \frac{3\pi}{4}+2k\pi$ với $k$ nguyên
Nếu $\cos x-2=0\Leftrightarrow \cos x=2$ (vô lý vì $\cos x\leq 1$)
b.
PT \(\Rightarrow \left[\begin{matrix} \tan x=\sqrt{3}\\ \tan x=1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{\pi}{3}+k\pi\\ x=\frac{\pi}{4}+k\pi\end{matrix}\right.\) với $k$ nguyên
c.
PT \(\Rightarrow \left[\begin{matrix} \cot \frac{x}{3}=1\\ \cot \frac{x}{2}=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{3}{4}\pi +3k\pi\\ x=\frac{-\pi}{2}+2k\pi \end{matrix}\right.\) với $k$ nguyên.
a/
\(\Leftrightarrow\left[{}\begin{matrix}2cosx+\sqrt{2}=0\\cosx-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{\sqrt{2}}{2}\\cosx=2>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\left[{}\begin{matrix}tanx-\sqrt{3}=0\\1-tanx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
c/ĐKXĐ: ...
\(\Leftrightarrow\left[{}\begin{matrix}cot\frac{x}{3}=1\\cot\frac{x}{2}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+k\pi\\\frac{x}{2}=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+k3\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)