Giải phương trình: \(\cot x+\sin x\left(1+\tan x\tan\frac{x}{2}\right)=4\)
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12 tháng 9 2016
tham khảogiúp mình nhé: (tanx + cotx)^2 - (tanx + cotx) = 2? | Yahoo Hỏi & Đáp
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) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2021
a.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)
\(\Leftrightarrow1-sin^2x=0\)
\(\Leftrightarrow cos^2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2021
b.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
LT
21 tháng 9 2019
\(PT\Leftrightarrow2+\frac{1}{sinxcosx}-cotx=-2-sinx-cosxcotx-tanx\)
\(\Leftrightarrow\frac{1}{sinxcosx}-\frac{cosx}{sinx}=-2sinx-\frac{sinx}{cosx}\)
\(\Leftrightarrow1-cos^2x+2sin^2xcosx+sin^2x=0\)
\(\Leftrightarrow2sin^2x+2sin^2xcosx=0\)
\(\Leftrightarrow2sin^2x\left(1+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\cosx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2k\pi\\x=\pi+2k\pi\end{matrix}\right.\Leftrightarrow x=k\pi\)
LT
22 tháng 9 2019
Mình sửa lại câu trả lời
ĐK:\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(PT\Leftrightarrow2sin^2x\left(1+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loai\right)\\cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow x=\pi+2k\pi\left(loai\right)\)
Vậy phương trình vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2019
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2019
f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
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) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
b.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{3}cot\pi x=\frac{\pi}{6}+k\pi\)
\(\Leftrightarrow cot\pi x=\frac{1}{2}+3k\)
\(\Leftrightarrow\pi x=arccot\left(\frac{1}{2}+3k\right)+n\pi\)
\(\Leftrightarrow x=\frac{1}{\pi}arccot\left(\frac{1}{2}+3k\right)+n\)
c.
\(\Leftrightarrow\left[{}\begin{matrix}\pi tan3x=\frac{\pi}{6}+k2\pi\\\pi tan3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tan3x=\frac{1}{6}+2k\\tan3x=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}arctan\left(\frac{1}{6}+2k\right)+\frac{n2\pi}{3}\\x=\frac{1}{3}arctan\left(\frac{5}{6}+2k\right)+\frac{n2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
a/
\(\Leftrightarrow\frac{\pi}{2}sin\pi\left(x+1\right)=\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow sin\pi\left(x+1\right)=\frac{1}{2}+2k\)
Do \(-1\le sin\pi\left(x+1\right)\le1\Rightarrow k=0\)
\(\Rightarrow sin\pi\left(x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{6}+k2\pi\\\pi\left(x+1\right)=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{1}{6}+2k\\x+1=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{6}+2k\\x=-\frac{1}{6}+2k\end{matrix}\right.\)
ĐKXĐ: ....
Bạn học cách đặt ẩn phụ \(t=tan\frac{x}{2}\) chưa nhỉ? Rồi thì bài này ngắn, còn chưa thì hơi dài, để an toàn cứ coi như chưa học đi:
Ta có: \(tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{2sin\frac{x}{2}.cos\frac{x}{2}}{2cos^2\frac{x}{2}}=\frac{sinx}{cosx+1}\)
Thay vào pt:
\(cotx+sinx\left(1+\frac{sin^2x}{cosx\left(1+cosx\right)}\right)=4\)
\(\Leftrightarrow cotx+sinx\left(\frac{cosx+cos^2x+sin^2x}{cosx\left(1+cosx\right)}\right)-4=0\)
\(\Leftrightarrow cotx+\frac{sinx}{cosx}-4=0\)
\(\Leftrightarrow\frac{1}{tanx}+tanx-4=0\)
Đặt \(t=tanx\Rightarrow t^2-4t+1=0\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2+\sqrt{3}=tan\left(\frac{5\pi}{12}\right)\\tanx=2-\sqrt{3}=tan\left(\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)