2sin\(^2\)x + sin2x -s 2sinx -cosx - sinx = 0
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: \(\left\{{}\begin{matrix}cosx< >0\\sinx< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{2}+k\Omega\\x\ne k\Omega\end{matrix}\right.\)
=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)
\(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left[sinx\cdot\cos\dfrac{\Omega}{3}+sin\left(\dfrac{\Omega}{3}\right)\cdot cosx\right]\)
=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left(\dfrac{1}{2}\cdot sinx+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\)
=>\(\left(sinx+\sqrt{3}\cdot cosx\right)\left(\dfrac{1}{cosx\cdot sinx}-1\right)=0\)
=>\(2\cdot\left(sinx\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\cdot\left(\dfrac{2}{2\cdot sinx\cdot cosx}-1\right)=0\)
=>\(2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\cdot\left(\dfrac{2}{sin2x}-1\right)=0\)
=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{\Omega}{3}\right)=0\\\dfrac{2}{sin2x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=k\Omega\\sin2x=2\left(loại\right)\end{matrix}\right.\)
=>\(x=-\dfrac{\Omega}{3}+k\Omega\)
Tìm GTLN và GTNN của hàm số : 1. y = sinx + 2cosx +1 / 2sinx + cosx + 3
2.y= 2sin^2sinx - 3 sinx cosx + cos^2 x
Giải phương trình : 1. 2sin^2 * 2x + sin7x -1 = sinx
2.cos 4x + 12 sin^2 x -1 = 0
Giải pt
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(sin2x-cos2x+3sinx-cosx-1=0\)
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
1) 2sinx + cosx = sin2x + 1
2) (1 + cosx)(1+sinx) = 2
3) 3cos4x - 8cos6x + 2cos2x +3 =0
4) sin3x + cos3x.sinx + cosx = \(\sqrt{2}\)cos2x
5) (2cosx -1)(2sinx + cosx) = sin2x - sinx
giai pt
a) \(\sqrt{3}cosx-sinx=2sin4x\)
b) \(sin2x+4sinx.cos^2x=2sinx\)
c) \(sin7x-sinx=1-2sin^22x\)
d) \(\frac{2sinx+cosx+1}{sinx-2cosx+3}=\frac{1}{3}\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx=sin4x\)
\(\Leftrightarrow sin\left(\frac{\pi}{3}-x\right)=sin4x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-x+k2\pi\\4x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow2sinx.cosx+4sinx.cos^2x-2sinx=0\)
\(\Leftrightarrow2sinx\left(cosx+2cos^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\2cos^2x+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow2cos4x.sin3x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2sin3x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\3x=\frac{\pi}{6}+k2\pi\\3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow6sinx+3cosx+3=sinx-2cosx+3\)
\(\Leftrightarrow sinx+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)
Giaỉ các phương trình lượng giác sau:
1. 2sin2x+3sinx=3cosx
2. sin2x-4(sinx-cosx)=4
3. (1+sinx)(1+cosx)=2
4. 2(sinx-cosx)-sin2x-1=0
5. sinx-cosx+4sinxcosx+1=0
6. sinx=2cos\(^3\)x
7. cosx=2sin\(^3\)x
8. 2cos\(^3\)x=sin3x
1.
\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(2\left(1-t^2\right)+3t=0\)
\(\Leftrightarrow-2t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
2.
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t=4\)
\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sinx-cosx=-1\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{3\pi}{2}+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow1+cosx+sinx+sinx.cosx=2\)
\(\Leftrightarrow2\left(sinx+cosx\right)+2sinx.cosx-2=0\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)
Pt trở thành:
\(2t+t^2-1-2=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sinx+cosx=1\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
1> 1 + sinx + cosx + sin2x + cos2x = 0
2> cos2x + 3sin2x + 5 sinx - 3cosx = 3
3> \(\dfrac{\sqrt{2}*(cosx - sinx)}{cotx - 1}\) = \(\dfrac{1}{tanx + cot2x}\)
4> (2cosx - 1)*(2sinx + cosx) = sin2x - sinx
Nghiệm phương trình: cosx ( cosx + 2 sinx ) + 3 sinx ( sinx + 2 ) sin 2 x - 1 = 1
A. x = ± π 4 + k2π, k ∈ Z
B. x = - π 4 + kπ, k ∈ Z
C. x = - π 4 + k2π, x = - 3 π 4 + k2π, k ∈ Z
D. x = - π 4 + k2π, k ∈ Z
Giải phương trình sau: cosx ( cosx + 2 sinx ) + 3 sinx ( sinx + 2 ) sin 2 x - 1 = 1
A. x = - π 4 + k2π
B. x = - 3 π 4 + k2π
C. x = ± π 4 + kπ
D. Cả A và B đúng