giải phương trình:
\(\left(\cos2x+\sin2x\right)\cos x+2\cos2x-\sin x=0\)
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HT
3
11 tháng 4 2016
Phương trình đã cho tương đương với :
\(1+\frac{\sqrt{3}}{2}\sin2x-\frac{1}{2}\cos2x-3\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)=0\)
\(\Leftrightarrow1-\cos\left(2x+\frac{\pi}{3}\right)-3\sin\left(x+\frac{\pi}{6}\right)=0\)
\(2\sin^2\left(x+\frac{\pi}{6}\right)-2\sin\left(x+\frac{\pi}{6}\right)=0\Leftrightarrow\begin{cases}\sin\left(x+\frac{\pi}{6}\right)=0\\\sin\left(x+\frac{\pi}{6}\right)=\frac{3}{2}\end{cases}\) (Loại \(\sin\left(x+\frac{\pi}{6}\right)=\frac{3}{2}\))
Với \(\sin\left(x+\frac{\pi}{6}\right)=0\Rightarrow x=-\frac{\pi}{6}+k\pi,k\in Z\)
3 tháng 7 2021
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
3 tháng 7 2021
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
12 tháng 7 2018
√ 3sin 2x+cos 2x=2cosx-1
<=>2√3 sinx.cox+cos2x -sin2x -2cosx+cos2x+sin2x=0
<=>2√3sinx.cosx+2cos2x -2cosx=0
<=>cosx(√3sinx+cosx -1)=0
*cosx=0 =>x=pi/2+k.pi
*√3sinx+cosx -1=0
<=>sin(x+pi/6)=1/2 <=>x=...
TC
GPT: \(\dfrac{\left(\sin x-\cos x\right)\left(\sin2x-3\right)-\sin2x-\cos2x+1}{2\sin x-\sqrt{2}}=0\)
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 1 2021
ĐKXĐ: \(sinx\ne\dfrac{\sqrt{2}}{2}\)
\(\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\\left(sin2x-1\right)+2\left(sinx+1\right)=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{\pi}{4}+k2\pi\)
13 tháng 8 2023
a: ĐKXĐ: sin 2x<>1
=>2x<>pi/2+k2pi
=>x<>pi/4+kpi
\(\dfrac{cos2x}{sin2x-1}=0\)
=>cos2x=0
=>2x=pi/2+kpi
=>x=pi/4+kpi/2
Kết hợp ĐKXĐ, ta được:
x=3/4pi+k2pi hoặc x=7/4pi+k2pi
b: cos(sinx)=1
=>sin x=kpi
=>sin x=0
=>x=kpi
c: \(2\cdot sin^2x-1+cos3x=0\)
=>cos3x+cos2x=0
=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)
=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)
=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi
=>x=-pi+k2pi hoặc x=pi/5+k2pi/5
e: cos3x=-cos7x
=>cos3x=cos(pi-7x)
=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi
=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2
XG
0
\(\Leftrightarrow cos2x.cosx+2cos2x+sin2x.cosx-sinx=0\)
\(\Leftrightarrow cos2x.cosx+2cos2x+2sinx.cos^2x-sinx=0\)
\(\Leftrightarrow cos2x.cosx+2cos2x+sinx\left(2cos^2x-1\right)=0\)
\(\Leftrightarrow cos2x.cosx+2cos2x+sinx.cos2x=0\)
\(\Leftrightarrow cos2x\left(cosx+2+sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)