giai pt
a) \(sin^6x+cos^6x+sin2x=1\)
b) \(1+sinx+cosx+sin2x+cos2x=0\)
c) \(sin^2x-3cos^2x=2\left(sinx+\sqrt{3}cosx\right)\)
d) \(sin^3x.cosx-sinx.cos^3x=\frac{\sqrt{2}}{8}\)
giải các pt
a) \(sinx+cosx-\sqrt{2}sin2x=0\)
b) \(sin^2x+sin2x=3cos^2x\)
c) \(sinx\left(1-sinx\right)=cosx\left(cosx-1\right)\)
d) \(2\left(sin^3x-cos^3x\right)=\sqrt{3}.cos2x\left(sinx-cosx\right)\)
a/
\(\Leftrightarrow sinx+cosx=\sqrt{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}sin2x\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=sin2x\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{\pi}{4}+k2\pi\\2x=\frac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=\frac{\pi}{4}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow\frac{1-cos2x}{2}+sin2x=\frac{3\left(1+cos2x\right)}{2}\)
\(\Leftrightarrow sin2x-2cos2x=1\)
\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\frac{1}{\sqrt{5}}\)
Đặt \(\frac{1}{\sqrt{5}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Leftrightarrow sin2x.cosa-cos2a.sina=cosa\)
\(\Leftrightarrow sin\left(2x-a\right)=cosa=sin\left(\frac{\pi}{2}-a\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-a=\frac{\pi}{2}-a+k2\pi\\2x-a=a-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=a-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sinx-sin^2x=cosx-cos^2x\)
\(\Leftrightarrow sinx-cosx-\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow sinx-cosx-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(1-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\1-sinx-cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\1-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)
\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)
\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)
giai pt:
a) \(\left(2cosx-1\right)\left(2sinx+cosx\right)=sin2x-sinx\)
b) \(\frac{sin2x}{cosx}+\frac{cos2x}{sinx}=tanx-cotx\)
c) \(\frac{1}{cos^2x}=\frac{2-sin^3x-cos^2x}{1-sin^3x}\)
a/
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)
\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)
\(\Leftrightarrow cosx=sin^2x-cos^2x\)
\(\Leftrightarrow cosx=1-2cos^2x\)
\(\Leftrightarrow2cos^2x+cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
1,Giải phương trình:
a,\(cos^3x+sin^3x=cos2x\)
b,\(cos^3x+sin^3x=2sin2x+sinx+cosx\)
c,\(2cos^3x=sin3x\)
d,\(cos^2x-\sqrt{3}sin2x=1+sin^2x\)
e,\(cos^3x+sin^3x=2\left(cos^5x+sin^5x\right)\)
a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)
b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x
⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x
⇔ 4sin2x + (sinx + cosx) . sin2x = 0
⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)
⇔ sin2x = 0
c, 2cos3x = sin3x
⇔ 2cos3x = 3sinx - 4sin3x
⇔ 4sin3x + 2cos3x - 3sinx(sin2x + cos2x) = 0
⇔ sin3x + 2cos3x - 3sinx.cos2x = 0
Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình
Xét cosx ≠ 0 chia cả 2 vế cho cos3x ta được :
tan3x + 2 - 3tanx = 0
⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
d, cos2x - \(\sqrt{3}sin2x\) = 1 + sin2x
⇔ cos2x - sin2x - \(\sqrt{3}sin2x\) = 1
⇔ cos2x - \(\sqrt{3}sin2x\) = 1
⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)
⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)
e, cos3x + sin3x = 2cos5x + 2sin5x
⇔ cos3x (1 - 2cos2x) + sin3x (1 - 2sin2x) = 0
⇔ cos3x . (- cos2x) + sin3x . cos2x = 0
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
giải các pt
a) \(sin2x-2\sqrt{3}cos^2x=4cosx\)
b) \(sin^2x-3cos^2x=sinx-\sqrt{3}cosx\)
c) \(sin6x\left(cos3x-1\right)-sin6x.sin3x=0\)
d) \(\left(sin2x-cos2x\right)^2-3\left(sin2x-cos2x\right)-4=0\)
e) \(\frac{sin2x+sin6x}{cos2x}-2cos4x=2\sqrt{2}\)
a/
\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x-4cosx=0\)
\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx-\sqrt{3}cosx=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=1\)
\(\Leftrightarrow x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{5\pi}{6}+k2\pi\)
b/
\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)=sinx-\sqrt{3}cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{3}cosx\left(1\right)\\sinx+\sqrt{3}cosx=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
\(\left(2\right)\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sin6x\left(cos3x-1-sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\Rightarrow x=\frac{k\pi}{6}\\cos3x-sin3x=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin3x-cos3x=-1\)
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\3x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{3}\\x=\frac{\pi}{2}+\frac{k2\pi}{3}\end{matrix}\right.\)
giải phương trình
1.\(sin^3x+2cosx-2+sin^2x=0\)
\(2.\frac{\sqrt{3}}{2}sin2x+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
3.\(2sin2x-cos2x=7sinx+2cosx-4\)
4.\(2cos2x-8cosx+7=\frac{1}{cosx}\)
5.\(cos^8x+sin^8x=2\left(cos^{10}x+sin^{10}x\right)+\frac{5}{4}cos2x\)
6.\(1+sinx+cos3x=cosx+sin2x+cos2x\)
7.\(1+sinx+cosx+sin2x+cos2x=0\)
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
3.
\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)
Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm
giai pt:
a) \(4sin^5x.cosx-4cos^5x.sinx=sin^24x\)
b) \(4sin^2\frac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\frac{3\pi}{4}\right)\)
c) \(sin^2\left(x+\frac{\pi}{3}\right)+sinx+\sqrt{3}cosx=\frac{5}{4}\)
d) \(2sinx\left(1+cos2x\right)+sin2x=1+2cosx\)
e) \(sin^2x+4sinx.cosx+3cos^2x-sinx-3ccosx=0\)
a/
\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)
\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)
\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)
\(\Leftrightarrow-sin4x=sin^24x\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
b/
\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)
\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)
\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(sinx+4cosx=2+sin2x\)
\(\left(1-sin2x\right)\left(sinx+cosx\right)=cos2x\)
\(1+sinx+cosx+sin2x+cos2x=0\)
\(sinx+sin2x+sin3x=1+cosx+cos2x\)
\(sin^22x-cos^28x=sin\left(\dfrac{17\pi}{2}+10x\right)\)
giải các pt
a) \(1-2cos2x-\sqrt{3}sinx+cosx=0\)
b) \(cos2x+cos^2x-sinx.cosx=8\left(cosx-sinx\right)\)
c) \(sin^2x+3sinx.cosx-4cos^2x=4\left(sinx-cosx\right)\)
d) \(\frac{cos^3x-sin^3x}{2cosx+3sinx}=cos2x\)
a/
\(\Leftrightarrow1-2\left(2cos^2x-1\right)-\sqrt{3}sinx+cosx=0\)
\(\Leftrightarrow3-4cos^2x+cosx-\sqrt{3}sinx=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(4cosx+3\right)-\sqrt{3}sinx=0\)
\(\Leftrightarrow2sin^2\frac{x}{2}\left(4cosx+3\right)-2\sqrt{3}sin\frac{x}{2}cos\frac{x}{2}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\frac{x}{2}=0\Rightarrow x=k2\pi\\sin\frac{x}{2}\left(4cosx+3\right)-\sqrt{3}cos\frac{x}{2}=0\left(1\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow sin\frac{x}{2}\left(8cos^2\frac{x}{2}-1\right)-\sqrt{3}cos\frac{x}{2}=0\)
- Với \(\left\{{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-1\end{matrix}\right.\) \(\Rightarrow x=-\pi+k4\pi\) là 1 nghiệm
- Với \(cos\frac{x}{2}\ne0\) chia 2 vế cho \(cos^3\frac{x}{2}\)
\(tan\frac{x}{2}\left(8-1-tan^2\frac{x}{2}\right)-\sqrt{3}-\sqrt{3}tan^2\frac{x}{2}=0\)
\(\Leftrightarrow-tan^3\frac{x}{2}-\sqrt{3}tan^2\frac{x}{2}+7tan\frac{x}{2}-\sqrt{3}=0\)
Đặt \(tan\frac{x}{2}=t\)
\(\Rightarrow t^3+\sqrt{3}t^2-7t+\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\sqrt{3}\\t=-2-\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{3}+k\pi\\\frac{x}{2}=-\frac{5\pi}{12}+k\pi\\\frac{x}{2}=\frac{\pi}{12}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{5\pi}{6}+k2\pi\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow cos^2x-sin^2x+cos^2x-sinx.cosx=8\left(cosx-sinx\right)\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)+cosx\left(cosx-sinx\right)=8\left(cosx-sinx\right)\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(2cosx+sinx-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\left(1\right)\\2cosx+sinx=8\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x-\frac{\pi}{4}=k\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
Xét (2), theo điều kiện có nghiệm của pt lượng giác bậc nhất, \(2^2+1^2< 8^2\Rightarrow\left(2\right)\) vô nghiệm
c/
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx\right)=4\left(sinx-cosx\right)\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+4cosx-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\sinx+4cosx-4=0\left(2\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
Xét (2) \(\Leftrightarrow\frac{1}{\sqrt{17}}sinx+\frac{4}{\sqrt{17}}cosx=\frac{4}{\sqrt{17}}\)
Đặt \(\frac{4}{\sqrt{17}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow cosx.cosa+sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x-a\right)=cosa\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=k2\pi\end{matrix}\right.\)
Giải các phương trình sau:
a, \(\sqrt{2}\) sin \(\left(2x+\frac{\pi}{4}\right)\)=3sinx+cosx+2
b, 1+sinx+cosx+sin2x+cos2x=0
c, (2cosx-1)(2sinx+cosx)=sin2x-sinx
d, cos3x+cos2x-cosx-1=0
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)