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\(1+sin5x=2cos^2\frac{x}{2}\)
\(\Leftrightarrow sin5x=2cos^2\frac{x}{2}-1\)
\(\Leftrightarrow sin5x=cosx\)
\(\Leftrightarrow sin5x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{2}-x+k2\pi\\5x=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
Câu đầu đơn giản là ko dịch được \(cos^22\times x/2\) nghĩa là gì :)
\(sin5x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow sin5x=-\left(cos^2x-sin^2x\right)\)
\(\Leftrightarrow sin5x=-cos2x\)
\(\Leftrightarrow sin5x=sin\left(2x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=2x-\frac{\pi}{2}+k2\pi\\5x=\frac{3\pi}{2}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
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a: \(\Leftrightarrow2\cdot\sin3x\cdot\cos x-2\cos^2x=0\)
\(\Leftrightarrow\cos x\left(\sin3x-\cos x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\\sin3x=\cos x=\sin\left(\dfrac{\Pi}{2}-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\3x=\dfrac{\Pi}{2}-x+k2\Pi\\3x=\dfrac{\Pi}{2}+x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
b: \(\Leftrightarrow\sin x+\sin5x+\sin^2x=0\)
\(\Leftrightarrow\sin x=0\)
hay \(x=k\Pi\)
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sin3x + 1=2sin22x
<=> sin3x + 1 = 2\(\dfrac{1-cos4x}{2}\)
<=> sin3x + 1 = 1 - cos4x
<=> sin3x = -cos4x
<=> sin3x + cos4x = 0
<=> \(\dfrac{\sqrt{2}}{2}\)sin3x + \(\dfrac{\sqrt{2}}{2}\)cos4x = 0 (chia 2 vế cho \(\sqrt{2}\)).
<=> cos\(\dfrac{\pi}{4}\)sin3x + sin\(\dfrac{\pi}{4}\)cos4x = 0
<=> sin (3x+\(\dfrac{\pi}{4}\)) = 0
<=> sin(3x+\(\dfrac{\pi}{4}\)) = sin0
<=> \(\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=0+k2\pi\\3x+\dfrac{\pi}{4}=\pi-0+k2\pi\end{matrix}\right.\)(k\(\in\)Z)
<=>\(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(k\(\in\)Z)
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1.
\(D=R\backslash\left\{\dfrac{\pi}{6}+\dfrac{k\pi}{3}\right\}\) là miền đối xứng
\(f\left(-x\right)=\left(-x^3-x\right)tan\left(-3x\right)=\left(x^3+x\right)tan3x=f\left(x\right)\)
Hàm chẵn
2.
\(D=R\)
\(f\left(-x\right)=\left(-2x+1\right)sin\left(-5x\right)=\left(2x-1\right)sin5x\ne\pm f\left(x\right)\)
Hàm không chẵn không lẻ
3.
\(D=R\backslash\left\{\dfrac{\pi}{6}+\dfrac{k\pi}{3}\right\}\) là miền đối xứng
\(f\left(-x\right)=tan\left(-3x\right).sin\left(-5x\right)=-tan3x.\left(-sin5x\right)=tan3x.sin5x=f\left(x\right)\)
Hàm chẵn
4.
\(D=R\)
\(f\left(-x\right)=sin^2\left(-2x\right)+cos\left(-10x\right)=sin^22x+cos10x=f\left(x\right)\)
Hàm chẵn
5.
\(D=R\backslash\left\{k\pi\right\}\) là miền đối xứng
\(f\left(-x\right)=\dfrac{-x}{sin\left(-x\right)}=\dfrac{-x}{-sinx}=\dfrac{x}{sinx}=f\left(x\right)\)
Hàm chẵn
![](https://rs.olm.vn/images/avt/0.png?1311)
4.
ĐKXĐ: \(2cos^2x+sinx-1\ne0\)
\(\Leftrightarrow-2sin^2x+sinx+1\ne0\Rightarrow\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)
Khi đó pt tương đương:
\(\Leftrightarrow\frac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{6}=-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(loại\right)\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
3.
\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)
\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)
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a, \(cos\left(x-\dfrac{\pi}{3}\right)-sin\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(x-\dfrac{7\pi}{12}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow x-\dfrac{7\pi}{12}=\pm\dfrac{\pi}{4}+k2\pi\)
...
b, \(\sqrt{3}sin2x+2cos^2x=2sinx+1\)
\(\Leftrightarrow\sqrt{3}sin2x+2cos^2x-1=2sinx\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+k2\pi\\2x+\dfrac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
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a/
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow3tan^2x+8tanx+8\sqrt{3}-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k\pi\end{matrix}\right.\)
b/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(tan^2x+2tanx-2=\frac{1}{2}\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^2x+4tanx-5=0\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-5\right)+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sinx+1\right)\left(1-2sin^2x-1\right)=0\)
\(\Leftrightarrow sin^2x\left(sinx+1\right)=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)