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\(\frac{1+sin2a}{1-sin2a}=\frac{sin^2a+cos^2a+2sina.cosa}{sin^2a+cos^2a-2sina.cosa}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)^2}\)
\(=\frac{\left(\sqrt{2}cos\left(a-\frac{\pi}{4}\right)\right)^2}{\left(\sqrt{2}sin\left(a-\frac{\pi}{4}\right)\right)^2}=\frac{cos^2\left(a-\frac{\pi}{4}\right)}{sin^2\left(a-\frac{\pi}{4}\right)}=cot^2\left(a-\frac{\pi}{4}\right)\)
\(cos^4x+sin^4x=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow\left(cos^2x+sin^2x\right)^2-2\left(sinx.cosx\right)^2=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow sin4x+sin^22x=0\)
\(\Leftrightarrow2sin2x.cos2x+sin^22x=0\)
\(\Leftrightarrow sin2x\left(2cos2x+sin2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=0\Rightarrow x=\frac{k\pi}{2}\\2cos2x+sin2x=0\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x+\frac{2}{\sqrt{5}}cos2x=0\)
Đặt \(cosa=\frac{1}{\sqrt{5}}\) với \(a\in\left[0;\pi\right]\)
\(\Rightarrow sin2x.cosa+cos2x.sina=0\)
\(\Leftrightarrow sin\left(2x+a\right)=0\)
\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)
\(2\left(cos^2x-1\right)=sinx.cos3x\)
\(\Leftrightarrow-2sin^2x=sinx.cos3x\)
\(\Leftrightarrow sinx.cos3x+2sin^2x=0\)
\(\Leftrightarrow sinx\left(cos3x+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos3x+2sinx=0\left(1\right)\end{matrix}\right.\)
Bạn có ghi nhầm đề ko nhỉ, pt (1) dù giải được nhưng khá khó đấy, phải vận dụng công thức nhân 3 và nghiệm ko hề đẹp
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
\( a){\mathop{\rm sinx}\nolimits} + \cos x = \sqrt 2 \sin 5x\\ \Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .\sin 5x\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 5x\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = 5x + k2\pi \\ x + \dfrac{\pi }{4} = \pi - 5x + k2\pi \end{array} \right.\left( {k \in \mathbb {Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)
\( b)\sqrt 3 \sin 2x + \sin \left( {\dfrac{\pi }{2} + 2x} \right) = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \sin \dfrac{\pi }{2}\cos 2x + \cos \dfrac{\pi }{2}\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + 1.\cos 2x + 0.\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \cos 2x - 1 = 0\\ \Leftrightarrow 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} + 1 - 2{\sin ^2}x - 1 = 0\\ \Leftrightarrow \sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - si{n^2}x = 0\\ \Leftrightarrow {\mathop{\rm sinx}\nolimits} \left( {\sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\mathop{\rm sinx}\nolimits} = 0\\ \sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \sin \left( {\dfrac{\pi }{3} - x} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \dfrac{\pi }{3} - x = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{3} - k\pi \end{array} \right. \)
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