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Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
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: ...
\(\Leftrightarrow cot\left(x-\frac{\pi}{4}\right)=cot\left(-x\right)\)
\(\Leftrightarrow x-\frac{\pi}{4}=-x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}\)
a/
\(y=\frac{1}{sinx}+\frac{1}{cosx}\ge\frac{4}{sinx+cosx}=\frac{4}{\sqrt{2}sin\left(x+\frac{\pi}{4}\right)}\ge\frac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=2\sqrt{2}\) khi \(\left\{{}\begin{matrix}sinx=cosx\\sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
\(y_{max}\) không tồn tại (y dần tới dương vô cùng khi x gần tới 0 hoặc \(\frac{\pi}{2}\))
b/
\(y=\frac{1}{1-cosx}+\frac{1}{1+cosx}=\frac{1+cosx+1-cosx}{1-cos^2x}=\frac{2}{sin^2x}\)
Hàm số ko tồn tại cả min lẫn max ( \(0< y< \infty\))
c/
Do \(tan^2x\) ko tồn tại max (tiến tới vô cực) trên khoảng đã cho nên hàm ko tồn tại max
\(y=2+\frac{sin^4x+cos^4x}{\left(sinx.cosx\right)^2}+\frac{1}{sin^4x+cos^4x}\ge2+2\sqrt{\frac{sin^4x+cos^4x}{\frac{1}{4}sin^22x.\left(sin^4x+cos^4x\right)}}\)
\(y\ge2+\frac{4}{sin2x}\ge2+\frac{4}{1}=6\)
\(y_{min}=6\) khi \(\left\{{}\begin{matrix}sin2x=1\\sin^4x+cos^4x=sinx.cosx\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
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) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
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.
\(\Leftrightarrow\left[{}\begin{matrix}\pi tan3x=\frac{\pi}{6}+k2\pi\\\pi tan3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tan3x=\frac{1}{6}+2k\\tan3x=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}arctan\left(\frac{1}{6}+2k\right)+\frac{n2\pi}{3}\\x=\frac{1}{3}arctan\left(\frac{5}{6}+2k\right)+\frac{n2\pi}{3}\end{matrix}\right.\)
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 \(-1\le sin\pi\left(x+1\right)\le1\Rightarrow k=0\)
\(\Rightarrow sin\pi\left(x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{6}+k2\pi\\\pi\left(x+1\right)=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{1}{6}+2k\\x+1=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{6}+2k\\x=-\frac{1}{6}+2k\end{matrix}\right.\)
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) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(2\cos 3x + 5 = 3\)
\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
a) Ta có:
\(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)
b) Ta có:
\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)
\(\text{1) Đ}K:\left\{{}\begin{matrix}sinx\ne0\\1-sinx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne m\pi\\x\ne\frac{\pi}{2}+n2\pi\end{matrix}\right.\)
\(2\text{) }ĐK:\left\{{}\begin{matrix}cos\left(2x+\frac{\pi}{3}\right)\ne0\\sinx\ne0\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}2x+\frac{\pi}{3}\ne\frac{\pi}{2}+m\pi\\x\ne n\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{12}+\frac{m\pi}{2}\\x\ne n\pi\end{matrix}\right.\)
\(3\text{) }ĐK:\left\{{}\begin{matrix}\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}\ge0\\1+sin\left(2x-\frac{\pi}{2}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-3cos2x\ge0\\sin\left(2x-\frac{\pi}{2}\right)\ne-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}cos2x\le\frac{5}{3}\left(T/m\right)\\2x-\frac{\pi}{2}\ne\frac{3\pi}{2}+k2\pi\end{matrix}\right.\Leftrightarrow x\ne\pi+k\pi\)
\(4\text{) }ĐK:\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)\ne0\\cos\left(3x-\frac{\pi}{4}\right)\ne0\\tan\left(3x-\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\frac{\pi}{3}\ne a\pi\\3x-\frac{\pi}{4}\ne\frac{\pi}{2}+b\pi\\3x-\frac{\pi}{4}\ne c\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{4}+\frac{b\pi}{3}\\x\ne\frac{\pi}{12}+\frac{c\pi}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{12}+\frac{k\pi}{6}\end{matrix}\right.\)