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Đặt \(cosx-sinx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)
\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)
Pt trở thành:
\(t\left(1+\dfrac{1-t^2}{2}\right)+1=0\)
\(\Leftrightarrow t^3-3t-2=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=-1\end{matrix}\right.\)
\(\Rightarrow cosx-sinx=-1\)
\(\Leftrightarrow\sqrt[]{2}cos\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=cos\left(\dfrac{3\pi}{4}\right)\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow cos2x-cos8x-sin3x+cos5x-2sin5x.cos5x=0\)
\(\Leftrightarrow2sin5x.sin3x-sin3x+cos5x-2sin5x.cos5x=0\)
\(\Leftrightarrow sin3x\left(2sin5x-1\right)-cos5x\left(2sin5x-1\right)=0\)
\(\Leftrightarrow\left(sin3x-cos5x\right)\left(2sin5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos5x=sin3x=cos\left(\dfrac{\pi}{2}-3x\right)\\sin5x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=3x-\dfrac{\pi}{2}+k2\pi\\5x=\dfrac{\pi}{6}+k2\pi\\5x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
3.
\(\Leftrightarrow1+sinx=cosx-cos3x+2sinx.cosx+1-2sin^2x\)
\(\Leftrightarrow sinx=2sin2x.sinx+2sinx.cosx-2sin^2x\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\1=2sin2x+2cosx-2sinx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4sinx.cosx+2cosx-2sinx-1=0\)
\(\Leftrightarrow2cosx\left(2sinx+1\right)-\left(2sinx+1\right)=0\)
\(\Leftrightarrow\left(2cosx+1\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
ĐK: `x \ne kπ`
`cot(x-π/4)+cot(π/2-x)=0`
`<=>cot(x-π/4)=-cot(π/2-x)`
`<=>cot(x-π/4)=cot(x-π/2)`
`<=> x-π/4=x-π/2+kπ`
`<=>0x=-π/4+kπ` (VN)
Vậy PTVN.
\(2sin^2\dfrac{x}{2}=cos5x+1\)
\(\Leftrightarrow-cos5x=1-2.sin^2\dfrac{x}{2}\)
\(\Leftrightarrow-cos5x=cosx\)
\(\Leftrightarrow cos\left(5x\right)=cos\left(\pi-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\pi-x+k2\pi\\5x=-\pi+x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\) (k nguyên)
Vậy..
những câu hỏi không liên quan đến THCS thì bạn vào h để có thể được giải đáp tốt hơn
a/
\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x=\sqrt{3}-2sin5x\)
\(\Leftrightarrow sin2x+\sqrt{3}\left(cos2x+1\right)=\sqrt{3}-2sin5x\)
\(\Leftrightarrow sin2x+\sqrt{3}cos2x=-2sin5x\)
\(\Leftrightarrow\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x=-sin5x\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=sin\left(-5x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=-5x+k2\pi\\2x+\frac{\pi}{3}=\pi+5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{21}+\frac{k2\pi}{7}\\x=-\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow sinx+\sqrt{3}cosx=2sin3x+2sinx\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=-2sin3x\)
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=-sin3x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=sin\left(-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=-3x+k2\pi\\x-\frac{\pi}{3}=\pi+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)
Ta có: sinx/2-cosx/2=1/2
<=> (sinx/2-cosx/2)2=1/4
<=> 1- sinx= 1/4
<=> sinx = 3/4
=> cosx = căn7/4 hoặc cosx= -căn7/4
=> sin2x = 2sinx.cosx
=> sin2x = 3. căn7/8 hoặc sin2x=-3.căn7/8