cho phương trình \(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}+...+\dfrac{1}{sin2^{2018}x}=0\)
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1 tháng 8 2021
ĐKXĐ: x≠ \(k.\dfrac{\pi}{4}\) với k ∈ Z
Pt đã cho tương đương
\(\left\{{}\begin{matrix}sin4x.sin2x+sin4x.cosx=sin2x.cosx\\x\ne k\dfrac{\pi}{4}\end{matrix}\right.\)
Do x≠ \(k.\dfrac{\pi}{4}\) với k ∈ Z nên sin2x ≠ 0, chia cả 2 vế cho sin2x ta được
sin4x + 2cos2x.cosx = cosx
⇔ sin4x = cosx (1 - 2cos2x)
⇔ 4sinx.cosx.cos2x = cosx (1 - 2cos2x)
Do x≠ \(k.\dfrac{\pi}{4}\) với k ∈ Z nên cosx ≠ 0, chia cả 2 vế cho cosx ta được
4sinx.cos2x = 1 - 2cos2x
⇔ 4.sinx(1 - 2sin2x) = 1 - 2. (1- 2sin2x)
Đến đây tự giải kết hợp điều kiện nhé
HP
5 tháng 9 2021
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
HP
5 tháng 9 2021
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
15 tháng 6 2021
a1)\(\dfrac{sin110}{cos110}+\dfrac{cos20}{sin20}\)
\(=\dfrac{sin\left(180-70\right)}{cos\left(180-70\right)}+\dfrac{cos\left(90-70\right)}{sin\left(90-70\right)}\)
\(=\dfrac{sin70}{-cos70}+\dfrac{sin70}{cos70}=0\)
a2) \(sin^2x+sin^2\left(\dfrac{\pi}{3}-x\right)+sinx.sin\left(\dfrac{\pi}{3}-x\right)\)
\(=\dfrac{1}{2}\left(1-cos2x\right)+\dfrac{1}{2}\left[1-cos\left(\dfrac{2\pi}{3}-2x\right)\right]+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(\dfrac{\pi}{3}\right)\right]\)
\(=\dfrac{1}{2}-\dfrac{1}{2}.cos2x+\dfrac{1}{2}-\dfrac{1}{2}.cos\left(\dfrac{2\pi}{3}-2x\right)+\dfrac{1}{2}.cos\left(2x-\dfrac{\pi}{3}\right)-\dfrac{1}{4}\)
\(=\dfrac{3}{4}-\dfrac{1}{2}\left[cos2x+cos\left(\dfrac{2\pi}{3}-2x\right)-cos\left(2x-\dfrac{\pi}{3}\right)\right]\)
\(=\dfrac{3}{4}-\dfrac{1}{2}\left[cos2x-2.sin\dfrac{\pi}{6}.sin\left(\dfrac{\pi-4x}{2}\right)\right]\)
\(=\dfrac{3}{4}-\dfrac{1}{2}\left(cos2x-cos2x\right)\)
\(=\dfrac{3}{4}\)
a3) \(sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(=\dfrac{1-cos2x}{2}+\dfrac{1}{2}\left[cos\left(-2x\right)+cos\left(\dfrac{2\pi}{3}\right)\right]\)
\(=\dfrac{1-cos2x}{2}+\dfrac{cos2x}{2}-\dfrac{1}{4}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\)
\(=\dfrac{1}{4}\)
HP
23 tháng 5 2021
b, \(VT=\dfrac{1-sin2x}{1+sin2x}\)
\(=\dfrac{sin^2x+cos^2x-2sinx.cosx}{sin^2x+cos^2x+2sinx.cosx}\)
\(=\dfrac{\left(sinx-cosx\right)^2}{\left(sinx+cosx\right)^2}\)
\(=\dfrac{\left(\dfrac{sinx-cosx}{cosx}\right)^2}{\left(\dfrac{sinx+cosx}{cosx}\right)^2}\)
\(=\dfrac{\left(\dfrac{sinx}{cosx}-1\right)^2}{\left(\dfrac{sinx}{cosx}+1\right)^2}\)
\(=\dfrac{\left(tanx-tan\dfrac{\pi}{4}\right)^2}{\left(1+tanx.tan\dfrac{\pi}{4}\right)^2}\)
\(=tan^2\left(x-\dfrac{\pi}{4}\right)=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\)
23 tháng 12 2023
b:
ĐKXĐ: \(\left\{{}\begin{matrix}cosx< >0\\sinx< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{2}+k\Omega\\x\ne k\Omega\end{matrix}\right.\)
=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)
\(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left[sinx\cdot\cos\dfrac{\Omega}{3}+sin\left(\dfrac{\Omega}{3}\right)\cdot cosx\right]\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left(\dfrac{1}{2}\cdot sinx+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\)
=>\(\left(sinx+\sqrt{3}\cdot cosx\right)\left(\dfrac{1}{cosx\cdot sinx}-1\right)=0\)
=>\(2\cdot\left(sinx\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\cdot\left(\dfrac{2}{2\cdot sinx\cdot cosx}-1\right)=0\)
=>\(2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\cdot\left(\dfrac{2}{sin2x}-1\right)=0\)
=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{\Omega}{3}\right)=0\\\dfrac{2}{sin2x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=k\Omega\\sin2x=2\left(loại\right)\end{matrix}\right.\)
=>\(x=-\dfrac{\Omega}{3}+k\Omega\)
DN
a) \(\sin2x=\dfrac{-1}{2}\)
b) \(\sin\dfrac{x}{2}=-1\)
c) \(\sin3x=\dfrac{1}{3}\)
d) \(\sin4x=-\sin x\)
5
21 tháng 9 2021
\(a,\sin2x=\dfrac{-1}{2}\Leftrightarrow\sin2x=\sin\left(-\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}+k2\pi\\2x=\pi+\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
\(\dfrac{1}{sin2k}=\dfrac{sink}{sink.sin2k}=\dfrac{\left(sin2k-k\right)}{sink.sin2k}=\dfrac{sin2k.cosk-cos2k.sink}{sink.sin2k}\)
\(=\dfrac{cosk}{sink}-\dfrac{cos2k}{sin2k}=cotk-cot2k\)
Do đó pt tương đương:
\(cot\dfrac{x}{2}-cotx+cotx-cot2x+...+cot2^{2017}x-cot^{2018}x=0\)
\(\Leftrightarrow cot\dfrac{x}{2}-cot2^{2018}x=0\)
\(\Leftrightarrow\dfrac{x}{2}=2^{2018}x+k\pi\)
\(\Leftrightarrow...\)
@Nguyễn VIệt Lâm giúp em với