Giải các phương trình sau:
1) sin3x + sinx.cosx + cos3x = 1
2) cosx + \(\sqrt{\text{3}}\)sinx = 2cos2x
3) \(\sqrt{\text{3}}\) ( cosx - \(\sqrt{\text{3}}\) sinx ) = 4sin2x.cos3x
Giải pt
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(sinx-\sqrt{3}cosx=2sin5x\)
\(\sqrt{3}cos5x-2sin3xcos2x-sinx=0\)
\(sinx+cosxsin2x+\sqrt{3}cos3x=2\left(cos4x-sin^3x\right)\)
\(tanx-3cotx=4\left(sinx+\sqrt{3}cosx\right)\)
1.
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=3x+\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)
2.
\(sinx-\sqrt{3}cosx=2sin5\text{}x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
Giải phương trình lượng giác bậc nhất đối với sinx và cosx:
\(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)
\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\frac{\text{sinx + sin2x + sin3x }}{\text{cosx + cos2x + cos3x}}\)=\(\sqrt{3}\)
ĐKXĐ: ..
\(\frac{sin3x+sinx+sin2x}{cos3x+cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\sqrt{3}\)
\(\Leftrightarrow tan2x=\sqrt{3}\)
\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)
Giải các phương trình sau:
a/ sinx + cosx = \(2\sqrt{2}\)sinx.cosx
b/ 3sinx - \(\sqrt{3}\)cosx = 0
c/ tanx . sinx +cosx . cosx = sinx + cosx
a) Đặt \(sinx+cosx=t\left(\left|t\right|\le\sqrt{2}\right)\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
=> pt có dạng: \(t=\sqrt{2}\left(t^2-1\right)\Leftrightarrow\sqrt{2}t^2-t-\sqrt{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{-\sqrt{2}}{2}\\t=\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}sinx+cosx=\frac{-\sqrt{2}}{2}\\sinx+cosx=\sqrt{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}sin\left(x+\frac{\pi}{4}\right)=\frac{-1}{2}\\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x+\frac{\pi}{4}=\frac{-\pi}{6}+2k\pi\\x+\frac{\pi}{4}=\frac{7\pi}{6}+2k\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-5\pi}{12}+2k\pi\\x=\frac{11\pi}{12}+2k\pi\\x=\frac{\pi}{4}+2k\pi\end{cases}}\left(k\inℤ\right)}\)
III. Phương trình bậc nhất đối với sinx và cosx:
*Giải các phương trình bậc nhất đối với sinx và cosx sau đây:
(2.1)
1) \(2sinx-2cosx=\sqrt{2}\)
2) \(cosx-\sqrt{3}sinx=1\)
3) \(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)
4) \(cosx-sinx=1\)
5) \(2cosx+2sinx=\sqrt{6}\)
6) \(sin3x+\sqrt{3}cosx=\sqrt{2}\)
7) \(3sinx-2cosx=2\)
(2.3)
1) \(\left(sinx-1\right)\left(1+cosx\right)=cos^2x\)
2) \(sin\left(\dfrac{\pi}{2}+2x\right)+\sqrt{3}sin\left(\pi-2x\right)=1\)
3) \(\sqrt{2}\left(cos^4x-sin^4x\right)=cosx+sinx\)
4) \(sin2x+cos2x=\sqrt{2}sin3x\)
5) \(sinx=\sqrt{2}sin5x-cosx\)
6) \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
7) \(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
8) \(2sin^2x+\sqrt{3}sin2x=3\)
9) \(sin^4x+cos^4\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{4}\)
(2.3)
1) \(\dfrac{\sqrt{3}\left(1-cos2x\right)}{2sinx}=cosx\)
2) \(cotx-tanx=\dfrac{cosx-sinx}{sinx.cosx}\)
3) \(\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}=4\)
4) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\)
5) \(3cosx+4sinx+\dfrac{6}{3cosx+4sinx+1}=6\)
(2.4)
a) Tìm nghiệm \(x\in\left(\dfrac{2\pi}{5};\dfrac{6\pi}{7}\right)\) của phương trình \(cos7x-\sqrt{3}sin7x+\sqrt{2}=0\)
b) Tìm nghiệm \(x\in\left(0;\pi\right)\) của phương trình \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\dfrac{3\pi}{4}\right)\)
(2.5) Xác định tham số m để các phương trình sau đây có nghiệm:
a) \(mcosx-\left(m+1\right)sinx=m\)
b) \(\left(2m-1\right)sinx+\left(m-1\right)cosx=m-3\)
(2.6) Tìm GTLN, GTNN (nếu có) của các hàm số sau đây:
a) \(y=3sinx-4cosx+5\)
b) \(y=cos2x+sin2x-1\)
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c.
\(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)
Câu này đề đúng không nhỉ? Nhìn thấy có vẻ không đúng lắm
d.
\(cosx-sinx=1\)
\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
Giải phương trình:lớp 11
1).\(\sin^2\left(x+\frac{\pi}{4}\right)=\sqrt{2}s\text{inx}\text{ }\text{ }\)
2)\(3\sqrt{2}cosx-sinx=cos3x+3\sqrt{2}sinxsin2x\)
sinx + sin2x + sin3x = 1 + cosx + cos2x
cos3x + sin3x + cosx - sinx = \(\sqrt{2}\)cos2x
sinx + sin2x + sin3x = cosx + cos2x + cos3x
b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)
c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
Giải phương trình:
a, sin2x+2sinx-cosx+1=0
b, \(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2sin\)(x+\(\dfrac{\text{π}}{3}\))
b:
ĐKXĐ: \(\left\{{}\begin{matrix}cosx< >0\\sinx< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{2}+k\Omega\\x\ne k\Omega\end{matrix}\right.\)
=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)
\(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left[sinx\cdot\cos\dfrac{\Omega}{3}+sin\left(\dfrac{\Omega}{3}\right)\cdot cosx\right]\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left(\dfrac{1}{2}\cdot sinx+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\)
=>\(\left(sinx+\sqrt{3}\cdot cosx\right)\left(\dfrac{1}{cosx\cdot sinx}-1\right)=0\)
=>\(2\cdot\left(sinx\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\cdot\left(\dfrac{2}{2\cdot sinx\cdot cosx}-1\right)=0\)
=>\(2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\cdot\left(\dfrac{2}{sin2x}-1\right)=0\)
=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{\Omega}{3}\right)=0\\\dfrac{2}{sin2x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=k\Omega\\sin2x=2\left(loại\right)\end{matrix}\right.\)
=>\(x=-\dfrac{\Omega}{3}+k\Omega\)
Giải phương trình: \(sin3x-cos3x+sinx+cosx=\dfrac{1}{sin3x+cosx}-\dfrac{1}{cos3x-sinx}\)
ĐKXĐ: ...
\(sin3x-cos3x+sinx+cosx=\dfrac{sin3x-cos3x+sinx+cosx}{\left(sin3x+cosx\right)\left(cos3x-sinx\right)}\)
\(\Rightarrow\left[{}\begin{matrix}sin3x-cos3x+sinx+cosx=0\left(1\right)\\\left(sin3x+cosx\right)\left(cos3x-sinx\right)=1\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow3sinx-4sin^3x-4cos^3x+3cosx+sinx+cosx=0\)
\(\Leftrightarrow sinx+cosx+sin^3x+cos^3x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)\left(1-sinx.cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2-sinx.cosx\right)=0\)
\(\Leftrightarrow sinx+cosx=0\) (loại)
(2) \(\Leftrightarrow sin3x.cos3x-sinx.cosx-sin3x.sinx+cos3x.cosx=1\)
\(\Leftrightarrow\dfrac{1}{2}sin6x-\dfrac{1}{2}sin2x+cos4x=1\)
\(\Leftrightarrow\dfrac{1}{2}\left(3sin2x-4sin^32x\right)-\dfrac{1}{2}sin2x+1-2sin^22x=1\)
\(\Leftrightarrow sin2x-2sin^32x-2sin^22x=0\)
\(\Leftrightarrow-sin2x\left(2sin^22x+2sin2x-1\right)=0\)
\(\Leftrightarrow...\)