GPT
a) \(tan\left(3x-\frac{\pi}{3}\right)=-tanx\)
b) \(cot\left(x-\frac{\pi}{4}\right)=-cotx\)
c) \(cot\left(2x-\frac{3\pi}{4}\right)=tan\left(x-\frac{\pi}{6}\right)\)
d) \(cos\left(2x+\frac{\pi}{3}\right)=-cos\left(x+\frac{\pi}{4}\right)\)
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
b.
\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)
\(\Leftrightarrow sinx-1=3k\)
\(\Leftrightarrow sinx=3k+1\)
Do \(-1\le sinx\le1\)
\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)
\(\Rightarrow k=0\)
\(\Rightarrow sinx=1\)
\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
c.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow cosx-1=4k-1\)
\(\Leftrightarrow cosx=4k\)
Mà \(-1\le cosx\le1\Rightarrow-1\le4k\le1\)
\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)
\(\Rightarrow cosx=0\)
\(\Rightarrow x=\frac{\pi}{2}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
16 tháng 7 2020
\(\text{1) Đ}K:\left\{{}\begin{matrix}sinx\ne0\\1-sinx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne m\pi\\x\ne\frac{\pi}{2}+n2\pi\end{matrix}\right.\)
\(2\text{) }ĐK:\left\{{}\begin{matrix}cos\left(2x+\frac{\pi}{3}\right)\ne0\\sinx\ne0\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}2x+\frac{\pi}{3}\ne\frac{\pi}{2}+m\pi\\x\ne n\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{12}+\frac{m\pi}{2}\\x\ne n\pi\end{matrix}\right.\)
\(3\text{) }ĐK:\left\{{}\begin{matrix}\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}\ge0\\1+sin\left(2x-\frac{\pi}{2}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-3cos2x\ge0\\sin\left(2x-\frac{\pi}{2}\right)\ne-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}cos2x\le\frac{5}{3}\left(T/m\right)\\2x-\frac{\pi}{2}\ne\frac{3\pi}{2}+k2\pi\end{matrix}\right.\Leftrightarrow x\ne\pi+k\pi\)
\(4\text{) }ĐK:\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)\ne0\\cos\left(3x-\frac{\pi}{4}\right)\ne0\\tan\left(3x-\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\frac{\pi}{3}\ne a\pi\\3x-\frac{\pi}{4}\ne\frac{\pi}{2}+b\pi\\3x-\frac{\pi}{4}\ne c\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{4}+\frac{b\pi}{3}\\x\ne\frac{\pi}{12}+\frac{c\pi}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{12}+\frac{k\pi}{6}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
c/ ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)
\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{12}+k\pi\\x=-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
Bạn tự tìm x thuộc khoảng đã cho
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
b/
ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow tan^22x+1+tan^22x=7\)
\(\Leftrightarrow tan^22x=3\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)
Bạn tự tìm nghiệm thuộc khoảng đã cho nhé
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 7 2020
a/
\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=tan\left(\frac{2\pi}{3}-3x\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{2\pi}{3}-3x+k\pi\)
\(\Rightarrow4x=\frac{\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{4}\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3}-\frac{3}{tanx}=0\)
\(\Leftrightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)
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
13 tháng 4 2019
\(sinx+cosx=\frac{1}{2}\Rightarrow\left(sinx+cosx\right)^2=\frac{1}{4}\Rightarrow sin^2x+cos^2x+2sinx.cosx=\frac{1}{4}\)
\(\Rightarrow2sinx.cosx=\frac{1}{4}-1=-\frac{3}{4}\Rightarrow sinx.cosx=-\frac{3}{8}\)
Vậy ta có:
\(sin^3x+cos^3x=\left(sinx+cosx\right)\left[\left(sinx+cosx\right)^2-3sinx.cosx\right]\)
\(=\frac{1}{2}\left(\frac{1}{4}+\frac{9}{8}\right)=\frac{11}{16}\)
Bài 2: Mục đích của bài này là gì bạn? Ko thấy yêu cầu?
Bài 3:
\(tanx+cotx=2\Rightarrow\left(tanx+cotx\right)^2=4\)
\(\Rightarrow tan^2x+2tanx.cotx+cot^2x=4\Rightarrow tan^2x+cot^2x+2=4\)
\(\Rightarrow tan^2x+cot^2x=2\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2019
\(=cos\left(4\pi+\pi+x\right)+sin\left(4\pi+\frac{\pi}{2}-x\right)-tan\left(\pi+\frac{\pi}{2}+x\right).cot\left(\pi+\frac{\pi}{2}-x\right)\)
\(=cos\left(\pi+x\right)+sin\left(\frac{\pi}{2}-x\right)-tan\left(\frac{\pi}{2}+x\right).cot\left(\frac{\pi}{2}-x\right)\)
\(=-cosx+cosx-\left(-cotx\right).tanx\)
\(=1\)
c.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(2x-\frac{3\pi}{4}\right)=cot\left(\frac{2\pi}{3}-x\right)\)
\(\Leftrightarrow2x-\frac{3\pi}{4}=\frac{2\pi}{3}-x+k\pi\)
\(\Leftrightarrow x=\frac{17\pi}{36}+\frac{k\pi}{3}\)
d.
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(\frac{3\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{3\pi}{4}-x+k2\pi\\2x+\frac{\pi}{3}=x-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{13\pi}{12}+k2\pi\end{matrix}\right.\)
a.
ĐKXĐ: ...
\(\Leftrightarrow tan\left(3x-\frac{\pi}{3}\right)=tan\left(-x\right)\)
\(\Leftrightarrow3x-\frac{\pi}{3}=-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{4}\)
b.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(x-\frac{\pi}{4}\right)=cot\left(-x\right)\)
\(\Leftrightarrow x-\frac{\pi}{4}=-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}\)