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Ta có:
\({\cos ^2}a + {\sin ^2}a = 1 \Rightarrow \sin a = \pm \frac{4}{5}\)
Do \(0 < a < \frac{\pi }{2} \Leftrightarrow \sin a = \frac{4}{5}\)
\(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{4}{3}\)
Ta có;
\(\begin{array}{l}\sin \left( {a + \frac{\pi }{6}} \right) = \sin a.\cos \frac{\pi }{6} + \cos a.\sin \frac{\pi }{6} = \frac{4}{5}.\frac{{\sqrt 3 }}{2} + \frac{3}{5}.\frac{1}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\cos \left( {a - \frac{\pi }{3}} \right) = \cos a.\cos \frac{\pi }{3} + \sin a.\sin \frac{\pi }{3} = \frac{3}{5}.\frac{1}{2} + \frac{4}{5}.\frac{{\sqrt 3 }}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\tan \left( {a + \frac{\pi }{4}} \right) = \frac{{\tan a + \tan \frac{\pi }{4}}}{{1 - \tan a.tan\frac{\pi }{4}}} = \frac{{\frac{4}{3} + 1}}{{1 - \frac{4}{3}}} = - 7\end{array}\)
\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)
\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
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:
\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)
a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)
Ta có: \({\sin ^2}a + {\cos ^2}a = 1\)
\(\Leftrightarrow \frac{1}{9} + {\cos ^2}a = 1\)
\(\Leftrightarrow {\cos ^2}a = 1 - \frac{1}{9}= \frac{8}{9}\)
\(\Leftrightarrow \cos a =\pm\sqrt { \frac{8}{9}} = \pm \frac{{2\sqrt 2 }}{3}\)
Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} = - \frac{{\sqrt 2 }}{4}\)
Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) = - \frac{{4\sqrt 2 }}{9}\)
\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)
\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} = - \frac{{4\sqrt 2 }}{7}\)
b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)
\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)
Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 = - \frac{3}{4}\)
Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)
\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)
\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)
\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)
\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)
\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 = - \frac{{\sqrt 7 }}{4}\)
\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)
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.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
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.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
c. ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{3}sin\pi x=\frac{\pi}{6}+k\pi\)
\(\Leftrightarrow sin\pi x=\frac{1}{2}+3k\)
\(-1\le\frac{1}{2}+3k\le1\Rightarrow k=0\)
\(\Rightarrow sin\pi x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi x=\frac{\pi}{6}+k2\pi\\\pi x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}+2k\\x=\frac{5}{6}+2k\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\pi}{6}cosx+\frac{\pi}{3}=k\pi\)
\(\Leftrightarrow cosx=-2+6k\)
Do \(-1\le cosx\le1\Rightarrow-1\le-2+6k\le1\)
\(\Rightarrow\frac{1}{6}\le k\le\frac{1}{2}\Rightarrow\) ko tồn tại k thỏa mãn
Vậy pt vô nghiệm
b.
\(\Leftrightarrow\pi cos3x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow cos3x=\frac{1}{2}+k\)
\(-1\le\frac{1}{2}+k\le1\Rightarrow k=\left\{-1;0\right\}\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
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
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
a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\). Do đó \(\cos a = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \frac{1}{3}} = - \frac{{\sqrt 6 }}{3}\)
Ta có: \(\cos \left( {a + \frac{\pi }{6}} \right) = \cos a\cos \frac{\pi }{6} - \sin a\sin \frac{\pi }{6} = - \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 3 }}.\frac{1}{2} = - \frac{{\sqrt 3 + 3\sqrt 2 }}{6}\)
b) Vì \(\pi < a < \frac{{3\pi }}{2}\) nên \(\sin a < 0\). Do đó \(\sin a = \sqrt {1 - {{\cos }^2}a} = \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a\; = \frac{{\sin a}}{{\cos a}} = \frac{{ - \frac{{2\sqrt 2 }}{3}}}{{ - \frac{1}{3}}} = 2\sqrt 2 \)
Ta có: \(\tan \left( {a - \frac{\pi }{4}} \right) = \frac{{\tan a - \tan \frac{\pi }{4}}}{{1 + \tan a\tan \frac{\pi }{4}}} = \frac{{\frac{{\sin a}}{{\cos a}} - 1}}{{1 + \frac{{\sin a}}{{\cos a}}}} = \frac{{2\sqrt 2 - 1}}{{1 + 2\sqrt 2 }} = \frac{{9 - 4\sqrt 2 }}{7}\)