Giải pt
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(sinx-\sqrt{3}cosx=2sin5x\)
\(\sqrt{3}cos5x-2sin3xcos2x-sinx=0\)
\(sinx+cosxsin2x+\sqrt{3}cos3x=2\left(cos4x-sin^3x\right)\)
\(tanx-3cotx=4\left(sinx+\sqrt{3}cosx\right)\)
\(\dfrac{\sqrt{2}\left(sinx-cox\right)^2\left(1+2sin2x\right)}{sin3x+sin5x}=1-tanx\)
\(sin\left(2x-\dfrac{\pi}{4}\right)cos2x-2\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
(sin2x+cos2x)cosx+2cos2x -sinx=0
sinx + cosxsin2x + \(\sqrt{3}cos3x=2\left(cos4x+sin^3x\right)\)
\(\sqrt{3}cos5x-2sin3xcos2x-sinx=0\)
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
\(sinx+cosx\cdot sin2x+\sqrt{3}cos3x=2.\left(cos4x+sin^3x\right)\)
\(sinx+cosx\cdot sin2x+\sqrt{3}cos3x=2.\left(cos4x+sin^3x\right)\)
\(\Leftrightarrow sinx+cosx\cdot sin2x+\sqrt{3}cos3x=2cos4x+2sin^3x\)
\(\Leftrightarrow sinx-2sin^3x+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sinx.\left(1-2sin^2x\right)+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sinx.cos2x+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sin.\left(x+2x\right)+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sin3x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow\dfrac{1}{2}sin3x+\dfrac{\sqrt{3}}{2}cos3x=cos4x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}.sin3x+sin\dfrac{\pi}{3}.cos3x=cos4x\)
\(\Leftrightarrow sin.\left(3x+\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{x}-4x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-4x+k2\pi\\3x+\dfrac{\pi}{2}=\pi-\dfrac{\pi}{2}+4x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{42}+\dfrac{k2\pi}{7}\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
Giải phương trình lượng giác bậc nhất đối với sinx và cosx:
\(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)
\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
giải pt
a) \(cosx\left(3tanx-\sqrt{3}\right)=0\)
b) \(\frac{\left(2-sinx\right)\left(\sqrt{3}cosx-1\right)}{1+sinx}+2=sinx\)
c) \(\frac{tanx-sinx}{sin^3x}=\frac{1}{cosx}\)
d) \(\frac{sin3x.cosx-sinx.cos3x}{cos^2x}=2\sqrt{3}\)
a/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow3tanx-\sqrt{3}=0\)
\(\Rightarrow tanx=\frac{1}{\sqrt{3}}\)
\(\Rightarrow x=\frac{\pi}{6}+k\pi\)
b/
ĐKXĐ: \(sinx\ne-1\)
\(\Leftrightarrow\frac{\left(2-sinx\right)\left(\sqrt{3}cosx-1\right)}{1+sinx}+2-sinx=0\)
\(\Leftrightarrow\left(2-sinx\right)\left(\frac{\sqrt{3}cosx-1}{1+sinx}+1\right)=0\)
\(\Leftrightarrow\frac{\sqrt{3}cosx-1}{1+sinx}=-1\) (do 2-sinx>0 với mọi x)
\(\Leftrightarrow\sqrt{3}cosx-1=-1-sinx\)
\(\Leftrightarrow sinx=-\sqrt{3}cosx\Rightarrow tanx=-\sqrt{3}\)
\(\Rightarrow x=-\frac{\pi}{3}+k\pi\)
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
Bài 1 Giải PT
a) sin3x - \(\sqrt{3}cos3x\) = 1
b) 3sin3x + \(\sqrt{3}cos9x\) = 1 + 4sin33x
c) \(\sqrt{3}cos4x\) + sin4x = 2
d) cos3x - sin2x = \(\sqrt{3}\)(cos3x - sin3x)
Bài 2: Cho PT 2m(sinx + cosx) = 2m2 + cosx - sinx +\(\frac{3}{2}\)
a) Giải PT với m= 1
b) Tìm m để PT có nghiệm
Có b nào gipus mk với cần gấp gấp :)
giải các pt
a) \(sinx+\sqrt{3}cosx=2sin\left(x+\frac{\pi}{6}\right)\)
b) \(\sqrt{3}sinx+cosx=2sin\frac{\pi}{12}\)
c) \(cosx-\sqrt{3}sinx=2cos3x\)
d) \(sin3x-\sqrt{3}cos3x=2sin2x\)
a/
\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=3x+k2\pi\\x+\frac{\pi}{3}=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)
Giải các pt:
a) \(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
b) \(2cos^2x-3\sqrt{3}sin2x-4sin^2x=-4\)
c) \(\sqrt{3}\left(cos2x+sin3x\right)=sin2x+cos8x\)
d) \(cos2x-\sqrt{3}sin2x=\sqrt{3}sinx+cosx\)
e) \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x+\frac{1}{2}cos3x=\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\)
\(\Leftrightarrow sin\left(3x+\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\3x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
b/
\(\Leftrightarrow2\left(\frac{1+cos2x}{2}\right)-3\sqrt{3}sin2x-4\left(\frac{1-cos2x}{2}\right)=-4\)
\(\Leftrightarrow3cos2x-3\sqrt{3}sin2x=-3\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)
\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{3}+k\pi\)
c/
Ủa đề câu này bạn ghi đúng ko? Nhìn kì kì, cos8x hay cos3x bên vế phải vậy?
d/
\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=x-\frac{\pi}{3}+k2\pi\\2x+\frac{\pi}{3}=\frac{\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2\pi}{3}+k2\pi\\x=\frac{k2\pi}{3}\end{matrix}\right.\)
e/
\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)
\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)
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/
\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)
\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)
\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)