( sinx + cosx)2 + 2sin2 x/2 = sinx (2sqrt3 sinx + 4 - sqrt3 )
giúo mình với!!!!
(sinx+cox)2 + 2sin2\(\dfrac{x}{2}\)=sinx( 2\(\sqrt{3}\)sinx +4-\(\sqrt{3}\))
\(\Leftrightarrow1+2sinx.cosx+1-cosx=2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx-2\right)=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2sinx-1\right)\left(\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(cosx+\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
Cmr:
1) (Sinx)/(1+cosx)+(1+cosx)/sinx=2/sinx
2) cosx/(1-sinx)=cot(bi/4-x/2)
\(\frac{sinx}{1+cosx}+\frac{1+cosx}{sinx}=\frac{sin^2x+\left(1+cosx\right)^2}{sinx\left(1+cosx\right)}=\frac{sin^2x+cos^2x+2cosx+1}{sinx\left(1+cosx\right)}\)
\(=\frac{2+2cosx}{sinx\left(1+cosx\right)}=\frac{2\left(1+cosx\right)}{sinx\left(1+cosx\right)}=\frac{2}{sinx}\)
\(\frac{cosx}{1-sinx}=\frac{cos2.\frac{x}{2}}{1-sin2.\frac{x}{2}}=\frac{cos^2\frac{x}{2}-sin^2\frac{x}{2}}{sin^2\frac{x}{2}+cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}=\frac{\left(cos\frac{x}{2}-sin\frac{x}{2}\right)\left(cos\frac{x}{2}+sin\frac{x}{2}\right)}{\left(cos\frac{x}{2}-sin\frac{x}{2}\right)^2}\)
\(=\frac{sin\frac{x}{2}+cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}}=\frac{\sqrt{2}cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{\sqrt{2}sin\left(\frac{\pi}{4}-\frac{x}{2}\right)}=cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
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.
\(cosx-sinx=1\)
\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
cho góc nhọn x biết
1. sinx=1/4 tính cosx. 2. tgx= 1/3, tính sinx.
mọi người giúp mình với ạ
1.
\(sin^2x+cos^2x=1\Rightarrow\left(\dfrac{1}{4}\right)^2+cos^2x=1\)
\(\Rightarrow cos^2x=\dfrac{15}{16}\Rightarrow cosx=\dfrac{\sqrt{15}}{4}\)
2.
\(tanx=\dfrac{1}{3}\Rightarrow tan^2x=\dfrac{1}{9}\Rightarrow\dfrac{sin^2x}{cos^2x}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{sin^2x}{1-sin^2x}=\dfrac{1}{9}\Rightarrow9sin^2x=1-sin^2x\)
\(\Rightarrow sin^2x=\dfrac{1}{10}\Rightarrow sinx=\dfrac{\sqrt{10}}{10}\)
chứng minh: cosx/sinx-cosx + sinx/sinx+cosx=1+cot2x/1-cot2x
giải phương trình
(sinx + cosx)^2 + 2sin^2 x/2 = sinx (2căn3 sinx +4 - căn3 )
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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
cosx- √2 sinx/sinx-1/2=0
mn giúp mk với ạ mình cần gấp ạ
Rút gọn :
A. Cot2x - cos2x = cos2x . Cot2x
B. Tanx + cosx / 1 + sinx = 1 / cosx
C. 2 / sinx - sinx / 1 + cosx = 1 + cosx / sinx