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4 tháng 9 2018

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)

14 tháng 6 2020

\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

NV
21 tháng 9 2020

\(A=\frac{cos3x+cos9x+cos5x+cos7x}{sin3x+sin9x+sin5x+sin7x}=\frac{2cos6x.cos3x+2cos6x.cosx}{2sin6x.cos3x+2sin6x.cosx}\)

\(=\frac{2cos6x\left(cos3x+cosx\right)}{2sin6x\left(cos3x+cosx\right)}=tan6x\)

\(A=1\Rightarrow tan6x=1\Rightarrow x=\frac{\pi}{24}+\frac{k\pi}{6}\)

23 tháng 9 2020

bằng cot6x chứ bạn???

NV
25 tháng 4 2019

\(\frac{sinx+sin5x+sin3x}{cosx+cos5x+cos3x}=\frac{2sin3x.cos2x+sin3x}{2cos3x.cos2x+cos3x}=\frac{sin3x\left(2cos2x+1\right)}{cos3x\left(2cos2x+1\right)}=\frac{sin3x}{cos3x}=tan3x\)

NV
20 tháng 9 2020

c/

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=1-cos4x\)

\(\Leftrightarrow cos6x+cos2x-2cos4x=0\)

\(\Leftrightarrow2cos4x.cos2x-2cos4x=0\)

\(\Leftrightarrow2cos4x\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)

NV
20 tháng 9 2020

a/

\(\Leftrightarrow1+cos2x+cos3x+cosx=0\)

\(\Leftrightarrow2cos^2x+2cos2x.cosx=0\)

\(\Leftrightarrow2cosx\left(cosx+cos2x\right)=0\)

\(\Leftrightarrow2cosx\left(2cos^2x+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

b/

\(\Leftrightarrow2sin3x.cosx+sin3x=2cos3x.cosx+cos3x\)

\(\Leftrightarrow sin3x\left(2cosx+1\right)-cos3x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left(sin3x-cos3x\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(3x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

NV
31 tháng 7 2020

c/

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)

d/

\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)

\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=cosx\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

NV
31 tháng 7 2020

a/

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)

\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)

\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)

b/

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)