giải các pt
a) \(\frac{1}{sin^2x}=cotx+3\)
b) \(\frac{\sqrt{3}}{sin^2x}=3cotx+\sqrt{3}\)
c) \(9-13cosx+\frac{4}{1+tan^2x}=0\)
d) \(2tan^2x+3=\frac{3}{cosx}\)
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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{1}{cos^2x}\left(9-13cosx\right)+4=0\)
\(\Leftrightarrow\frac{9}{cos^2x}-\frac{13}{cosx}+4=0\)
Đặt \(\frac{1}{cosx}=t\)
\(\Rightarrow9t^2-13t+4=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{4}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{1}{cosx}=\frac{4}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
d/
\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)
\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^22x+sin2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
10. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(2cos2x+tanx=\frac{4}{5}\)
\(\Leftrightarrow4cos^2x-2+tanx=\frac{4}{5}\)
\(\Leftrightarrow\frac{4}{1+tan^2x}+tanx-\frac{14}{5}=0\)
Đặt \(tanx=t\)
\(\Rightarrow\frac{20}{1+t^2}+5t-14=0\)
\(\Leftrightarrow5t^3-14t^2+5t+6=0\)
\(\Leftrightarrow\left(t-2\right)\left(5t^2-4t-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{2+\sqrt{19}}{5}\\t=\frac{2-\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2=tana\\tanx=\frac{2+\sqrt{19}}{5}=tanb\\tanx=\frac{2-\sqrt{19}}{5}=tanc\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\\x=c+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
9.
\(\Leftrightarrow cos2x-3cosx=2\left(cosx+1\right)\)
\(\Leftrightarrow2cos^2x-1-3cosx=2cosx+2\)
\(\Leftrightarrow2cos^2x-5cosx-3=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 7 2020
b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2020
a.
\(\Leftrightarrow2sin\frac{17\pi}{30}cos\left(3x-\frac{7\pi}{30}\right)=\sqrt{3}\)
\(\Leftrightarrow cos\left(3x-\frac{7\pi}{30}\right)=\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}\)
Đặt \(\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow cos\left(3x-\frac{7\pi}{30}\right)=cosa\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{7\pi}{30}=a+k2\pi\\3x-\frac{7\pi}{30}=-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{90}+\frac{a}{3}+\frac{k2\pi}{3}\\x=\frac{7\pi}{30}-\frac{a}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
Chắc bạn ghi sai đề, con số \(\frac{4\pi}{3}\) sẽ hợp lý hơn con số \(\frac{4\pi}{5}\) rất nhiều
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
19 tháng 8 2020
3.
ĐKXĐ; ..
\(\sqrt{3}tanx+\frac{1}{tanx}-\sqrt{3}-1=0\)
\(\Leftrightarrow\sqrt{3}tan^2x-\left(\sqrt{3}+1\right)tanx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\frac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)
4.
\(\Leftrightarrow2cos^2x-1-3cosx=2+2cosx\)
\(\Leftrightarrow2cos^2x-5cosx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=3>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 8 2020
1.
\(\Leftrightarrow3\left(2cos^22x-1\right)-\left(1-cos^22x\right)+cos2x-2=0\)
\(\Leftrightarrow7cos^22x+cos2x-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\frac{6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{1}{2}arccos\left(\frac{6}{7}\right)+k\pi\end{matrix}\right.\)
2.
ĐKXĐ: ...
\(\Leftrightarrow1+cot^2x+3cotx+1=0\)
\(\Leftrightarrow cot^2x+3cotx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot\left(-2\right)+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
e/
\(\Leftrightarrow\left(sin^2x+4sinx.cosx+3cos^2x\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+3cosx\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+3cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+3cosx=0\\sinx+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-3cosx\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-3\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-3\right)+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
d/
\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(2-2sin^2x-2sinx.cosx-sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cos^2x-2sinx.cosx-sinx+cosx\right]=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cosx\left(cosx-sinx\right)+cosx-sinx\right]=0\)
\(\Leftrightarrow-\left(sinx-cosx\right)^2\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c/
ĐKXĐ: ...
\(\Leftrightarrow9-13cosx+4.cos^2x=0\)
\(\Leftrightarrow\left(cosx-1\right)\left(4cosx-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
d/
\(\Leftrightarrow2\left(tan^2x+1\right)+1=\frac{3}{cosx}\)
\(\Leftrightarrow\frac{2}{cos^2x}-\frac{3}{cosx}+1=0\)
\(\Leftrightarrow\left(\frac{1}{cosx}-1\right)\left(\frac{2}{cosx}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{2}{cosx}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
a/
ĐKXĐ: ..
\(\Leftrightarrow1+cot^2x=cotx+3\)
\(\Leftrightarrow cot^2x-cotx-2=0\)
\(\Rightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot\left(2\right)+k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow\sqrt{3}\left(1+cot^2x\right)=3cotx+\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}cot^2x-3cotx=0\)
\(\Rightarrow\left[{}\begin{matrix}cotx=0\\cotx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)