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:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
1: \(\Leftrightarrow4\cdot\dfrac{1+\cos2x}{2}-6\cdot\dfrac{1-\cos2x}{2}+5\sin2x-4=0\)
\(\Leftrightarrow2+2\cos2x-3+3\cos2x+5\sin2x-4=0\)
\(\Leftrightarrow5\sin2x+5\cos2x=5\)
\(\Leftrightarrow\cos2x+\sin2x=1\)
\(\Leftrightarrow\sqrt{2}\cdot\sin\left(2x+\dfrac{\Pi}{4}\right)=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\2x+\dfrac{\Pi}{4}=\dfrac{3\Pi}{4}+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
2: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)
\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)
\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
1. \(4\cos^2x-6\sin^2x+5\sin2x-4=0\)
\(\Leftrightarrow4\cos^2x-6\sin^2x+10\sin x\cos x-4\left(\cos^2x+\sin^2x\right)=0\)
\(\Leftrightarrow10\sin x\cos x-10\sin^2x=0\)
\(\Leftrightarrow10\sin x\left(\cos x-\sin x\right)=0\)
2. \(\sqrt{3}\cos^2x+2\sin x\cos x-\sqrt{3}\sin^2x-1=0\)
\(\Leftrightarrow\left(\sqrt{3}\cos^2x+\sin x\cos x\right)+\left(\sin x\cos x-\sqrt{3}\sin^2x\right)-1=0\)
\(\Leftrightarrow2\cos x\left(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)+2\sin x\left(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\right)-1=0\)
\(\Leftrightarrow2\cos x.\cos\left(\dfrac{\Pi}{6}-x\right)+2\sin x.\sin\left(\dfrac{\Pi}{6}-x\right)-1=0\)
\(\Leftrightarrow\cos\dfrac{\Pi}{6}+\cos\left(2x-\dfrac{\Pi}{6}\right)+\cos\left(2x-\dfrac{\Pi}{6}\right)-\cos\dfrac{\Pi}{6}-1=0\)
\(\Leftrightarrow\cos\left(2x-\dfrac{\Pi}{6}\right)=\dfrac{1}{2}\)
3. \(2\sin^22x-3\sin2x\cos2x+\cos^22x=2\)
\(\Leftrightarrow2\sin^22x-3\sin2x\cos2x+\cos^22x-2\left(\sin^22x+\cos^22x\right)=0\)
\(\Leftrightarrow3\sin2x\cos2x+\cos^22x=0\)
\(\Leftrightarrow\cos2x\left(3\sin2x+\cos2x\right)=0\)
-TH1: ...
- TH2: \(\cos2x=-3\sin2x\) mà \(\cos^22x+\sin^22x=1\) suy ra ...
4. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)
\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\cos x+\sin x=-1\)
1.
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)+sinx.cosx-1=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)-\left(1-sinx.cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx-1\right)\left(1-sinx.cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=1\\sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\\\frac{1}{2}sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin2x=2\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx=cos2x\)
\(\Leftrightarrow cos2x=cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\frac{\pi}{3}+k2\pi\\2x=\frac{\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
\(\Leftrightarrow\sqrt{3}cosx-3sinx=2sin5x-2sinx\)
\(\Leftrightarrow\sqrt{3}cosx-sinx=2sin5x\)
\(\Leftrightarrow-\left(\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx\right)=sin5x\)
\(\Leftrightarrow sin5x=-sin\left(x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{3}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{3}-x+k2\pi\\5x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
e.
\(3\left(1-sin^2x\right)-5sinx-1=0\)
\(\Leftrightarrow-3sin^2x-5sinx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{3}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
f.
\(2\left(2cos^2x-1\right)-cosx+7=0\)
\(\Leftrightarrow4cos^2x-cosx+5=0\)
Phương trình vô nghiệm
g.
\(\Leftrightarrow\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)=2\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=\sqrt{2}>1\)
Phương trình vô nghiệm
h.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
Theo điều kiện của pt lượng giác bậc nhất:
\(m^2+\left(3m+1\right)^2\ge\left(1-2m\right)^2\)
\(\Leftrightarrow10m^2+6m+1\ge4m^2-4m+1\)
\(\Leftrightarrow3m^2+5m\ge0\Rightarrow\left[{}\begin{matrix}m\ge0\\m\le-\frac{5}{3}\end{matrix}\right.\)
4.
\(\Leftrightarrow1-sin^2x-\left(m^2-3\right)sinx+2m^2-3=0\)
\(\Leftrightarrow-sin^2x-m^2sinx+2m^2+3sinx-2=0\)
\(\Leftrightarrow\left(-sin^2x+3sinx-2\right)+m^2\left(2-sinx\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2-sinx\right)+m^2\left(2-sinx\right)=0\)
\(\Leftrightarrow\left(2-sinx\right)\left(sinx-1+m^2\right)=0\)
\(\Leftrightarrow sinx=1-m^2\)
\(\Rightarrow-1\le1-m^2\le1\)
\(\Rightarrow m^2\le2\Rightarrow-\sqrt{2}\le m\le\sqrt{2}\)
1.
Bạn xem lại đề, \(sin^2x\left(\frac{x}{2}-\frac{\pi}{4}\right)\) là sao nhỉ?Có cả x trong lẫn ngoài ngoặc?
2.
ĐKXĐ: \(sinx\ne0\)
\(\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=1-cos^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
a, ta có 2x + π/3 = 3π/4 +k2π hoặc 2x + π/3 = -3π/4 + k2π
=> x= 5π/24 + kπ hoặc x= -13π/24 +kπ
b, đề sai phải ko
c, cos22x - sin22x - 2sinx -1=0
<=> -2sin22x -2sin2x =0
<=> sin2x=0 hoặc sin2x=-1
<=> x=kπ hoặc x= π/2 + kπ ; x=-π/4 +kπ hoặc x=5π/8 + kπ
d, cos5xcosπ/4 - sin5xsinπ/4 = -1/2
cos( 5x + π/4 ) = -1/2
<=> x=π/12 +k2π/5 hoặc x= -11π/60 + k2π/5
f,4x+π/3=3π/10 -x +k2π hoặc 4x+π/3 = x - 3π/10 +k2π
<=> x =-π/150 + k2π/5 hoặc x = π/90 +k2π/3
a)pt\(\Leftrightarrow cosx\left(cosx+1\right)+sinx.sin^2x=0\)
\(\Leftrightarrow cosx\left(cosx+1\right)+sinx\left(1-cos^2x\right)=0\)
\(\Leftrightarrow\left(cosx+1\right)\left(cosx+sinx-sinx.cosx\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}cosx=1\Leftrightarrow x=\pi+k2\pi\\cosx+sinx-sinx.cosx=0\left(\cdot\right)\end{array}\right.\)
Xét pt(*):
Đặt \(t=cosx+sinx,t\in\left[-\sqrt{2};\sqrt{2}\right]\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
(*) trở thành:\(t^2-2t-1=0\Leftrightarrow\left[\begin{array}{nghiempt}t=1-\sqrt{2}\\t=1+\sqrt{2}\left(L\right)\end{array}\right.\)
+)\(t=1-\sqrt{2}\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\\ \Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{\pi}{4}+arcsin\left(\frac{-2+\sqrt{2}}{2}\right)+k2\pi\\x=-\frac{5\pi}{4}-arcsin\left(\frac{-2+\sqrt{2}}{2}\right)+k2\pi\end{cases}\left(k\in Z\right)}\)
d/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow2\sqrt{2}\left(tanx+1\right)=\frac{3}{cos^2x}+2\)
\(\Leftrightarrow2\sqrt{2}tanx+2\sqrt{2}=3\left(1+tan^2x\right)+2\)
\(\Leftrightarrow3tan^2x-2\sqrt{2}tanx+5-2\sqrt{2}=0\)
Pt vô nghiệm
c/
\(\Leftrightarrow1-sin^2x+\sqrt{3}sinx.cosx-1=0\)
\(\Leftrightarrow\sqrt{3}sinx.cosx-sin^2x=0\)
\(\Leftrightarrow sinx\left(\sqrt{3}cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\\sqrt{3}cosx=sinx\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)