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a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
\(\Leftrightarrow2cosx.cos2x-\left(cosx+sinx\right)-\sqrt{2}sin3x\left(cosx+sinx\right)=0\)
\(\Leftrightarrow2cosx\left(cos^2x-sin^2x\right)-\left(cosx+sinx\right)\left(1+\sqrt{2}sin3x\right)=0\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(2cos^2x-2sinx.cosx\right)-\left(cosx+sinx\right)\left(1+\sqrt{2}sin3x\right)=0\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(2cos^2x-sin2x-1-\sqrt{2}sin3x\right)=0\)
Biến đổi ngoặc sau:
\(cos2x-sin2x=\sqrt{2}sin3x\)
\(\Leftrightarrow-\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin3x\)
\(\Leftrightarrow sin\left(\frac{\pi}{4}-2x\right)=sin3x\)
ta có \(\hept{\begin{cases}\sqrt{2}\left(sinx+cosx\right)=2sin\left(x+\frac{\pi}{4}\right)\\sinx.cosx=\frac{1}{2}sin2x=-\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)=-\frac{1-2sin^2\left(x+\frac{\pi}{3}\right)}{2}\end{cases}}\)
Vậy phương trình \(\Leftrightarrow2sin\left(x+\frac{\pi}{4}\right)+\frac{1-2sin^2\left(x+\frac{\pi}{4}\right)}{2}=1\)
Đặt \(sin\left(x+\frac{\pi}{4}\right)=a\Rightarrow PT\Leftrightarrow2a+\frac{1-2a^2}{2}=1\Leftrightarrow\orbr{\begin{cases}a=1+\frac{1}{\sqrt{2}}\\a=1-\frac{1}{\sqrt{2}}\end{cases}}\)
vì sin <1 nên \(sin\left(x+\frac{\pi}{4}\right)=1-\frac{1}{\sqrt{2}}\)có 4 nghiệm trên \(\left(0,2\pi\right)\)
a) Ta có : \(sin\left(x-\frac{2\pi}{3}\right)=cos2x\)
\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2\pi}{3}=\frac{\pi}{2}-2x+k2\pi\\x-\frac{2\pi}{3}=\pi-\frac{\pi}{2}+2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{18}+k\frac{2\pi}{3}\\x=-\frac{7\pi}{6}-k2\pi\end{matrix}\right.\)
Vậy ...
\(\Leftrightarrow sin^{2015x}-2sin^{2017}x-cos^{2016}x+2cos^{2018}x-cos2x=0\)
\(\Leftrightarrow sin^{2015}x\left(1-2sin^2x\right)+cos^{2016}x\left(2cos^2x-1\right)-cos2x=0\)
\(\Leftrightarrow cos2x\left(sin^{2015}x+cos^{2016}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin^{2015}x+cos^{2016}x=1\end{matrix}\right.\)
\(cos2x=0\Rightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\left\{{}\begin{matrix}sin^{2015}x\le sin^2x\\cos^{2016}x\le cos^2x\end{matrix}\right.\) \(\Rightarrow sin^{2015}x+cos^{2016}x\le sin^2x+cos^2x=1\)
Dấu "=" xảy ra khi và chỉ khi: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=0\\cosx=\pm1\end{matrix}\right.\\\left\{{}\begin{matrix}cosx=0\\sin=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(-10\le\frac{\pi}{4}+\frac{k\pi}{2}\le30\Rightarrow k=...\)
\(-10\le k\pi\le30\Rightarrow k=...\)
\(-10\le\frac{\pi}{2}+k2\pi\le30\Rightarrow k=...\)
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