Giải các phương trình sau:
1) \(2\cos4x-3=0\)
2) \(cos5x+2=0\)
3) \(cos2x+0,7=0\)
4) \(cos^22x-\dfrac{1}{4}=0\)
1) cos3x - cos4x + cos5x =0
2) sin3x + cos2x = 1 + 2sinx.cos2x
3) cos2x - cosx = 2 sin\(^2\)\(\dfrac{3x}{2}\)
4) cos\(^2\)2x + cos\(^2\)3x = sin\(^2\)x
5) sin3x.sin5x - cos4x.cos6x = 0
2.
\(sin3x+cos2x=1+2sinx.cos2x\)
\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)
\(\Leftrightarrow cos2x+sinx-1=0\)
\(\Leftrightarrow-2sin^2x+sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
1.
\(cos3x-cos4x+cos5x=0\)
\(\Leftrightarrow cos3x+cos5x-cos4x=0\)
\(\Leftrightarrow2cos4x.cosx-cos4x=0\)
\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)
3.
\(cos2x-cosx=2sin^2\dfrac{3x}{2}\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}+2sin^2\dfrac{3x}{2}=0\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}+sin\dfrac{3x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{3x}{2}.sinx.cos\dfrac{x}{2}=0\)
Đến đây dễ rồi tự làm tiếp nha.
Giải các phương trình sau:
\(a,cos2x+4cosx+1=0\)
\(b,cos^22x=\dfrac{1}{4}\)
a, \(cos2x+4cosx+1=0\)
\(\Leftrightarrow2cos^2x+4cosx=0\)
\(\Leftrightarrow2cosx\left(cosx+2\right)=0\)
\(\Leftrightarrow cosx=0\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
b, \(cos^22x=\dfrac{1}{4}\)
\(\Leftrightarrow4cos^22x-1=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow cosx=\pm\dfrac{1}{2}\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k\pi\)
Mọi người giúp mình bài này với (lớp 11)
1) \(\text{ 2cos4x*(cos2x-cos4x)=0 (mình ko biết giải phần cos2x-cos4x)}\)
2) \(\frac{1}{2}\left(cos5x-cos7x\right)=cos^22x-cos^23x\)
3) \(sin^22x=sin^23x\)
Đăng lên chô khác đi :D đây toàn lớp THCS có lẽ ít ai giải :v
vị dụ VMF , HMF, h,...................................><
Giải các phương trình lượng giác sau:
1) a/ \(cos\left(10x+12\right)+4\sqrt{2}sin\left(5x+6\right)-4=0\)
b/ \(cos\left(4x+2\right)+3sin\left(2x+1\right)=2\)
2) a/ \(cos2x+sin^2x+2cosx+1=0\)
b/ \(4sin^22x-8cos^2x+ 3=0\)
c/ \(4cos2x+4sin^2x+4sinx=1\)
3) a/ \(tanx+cotx=2\)
b/ \(2tanx-2cotx=3\)
4) a/ \(2sin2x+8tanx=9\sqrt{3}\)
b/ \(2cos2x+tan^2x=5\)
5) a/ \(\left(3+cotx\right)^2=5\left(3+cotx\right)\)
b/ \(4\left(sin^2x+\dfrac{1}{sin^2x}\right)-4\left(sinx+\dfrac{1}{sinx}\right)=7\)
1a.
Đặt \(5x+6=u\)
\(cos2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
1b.
Đặt \(2x+1=u\)
\(cos2u+3sinu=2\)
\(\Leftrightarrow1-2sin^2u+3sinu=2\)
\(\Leftrightarrow2sin^2u-3sinu+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
2a.
\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)
\(\Leftrightarrow cos^2x+2cosx+1=0\)
\(\Leftrightarrow\left(cosx+1\right)^2=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\)
Giải các phương trình sau :
a) \(2\cos x-\sin x=2\)
b) \(\sin5x+\cos5x=-1\)
c) \(8\cos^4x-4\cos2x+\sin4x-4=0\)
d) \(\sin^6x+\cos^6x+\dfrac{1}{2}\sin4x=0\)
giải phương trình
a) \(cos5x=-5\)
b) \(2cosx-1=0\)
c) \(-5cos\left(x+\dfrac{\pi}{3}\right)=0\)
d) \(cos4x=cos\dfrac{5\pi}{12}\)
e) \(cos^2x=1\)
a: cos5x=-5
mà -1<=cos5x<=1
nên \(x\in\varnothing\)
b: 2*cosx-1=0
=>2*cosx=1
=>cosx=1/2
=>x=pi/3+k2pi hoặc x=-pi/3+k2pi
c: -5*cos(x+pi/3)=0
=>cos(x+pi/3)=0
=>x+pi/3=pi/2+kpi
=>x=pi/6+kpi
d: cos4x=cos(5/12pi)
=>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
e: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
Giải phương trình sau: \(\sin^22x-2\cos x+\dfrac{3}{4}=0\)
Giải PT
a1) \(\dfrac{\left(1-2\sin x\right)\cos x}{\left(1+2\sin x\right)\left(1-\sin x\right)}=\sqrt{3}\)
a2) \(2\sin17x+\sqrt{3}\cos5x+\sin5x=0\)
a3) \(\)\(\cos7x-\sin5x=\sqrt{3}\left(\cos5x-\sin7x\right)\)
a4) \(\sqrt{3}\cos5x-2\sin3x\cos2x-\sin x=0\)
a5) \(\tan x+\cot x=2\left(\sin2x+\cos2x\right)\)
1. Cho biết \(cosx=\dfrac{3}{4}\). Tính giá trị của biểu thức \(P=sin^22x\).
2. Giải phương trình \(cos2x-sin\left(x+\dfrac{\pi}{3}\right)=0\)
1: \(P=sin^22x=1-cos^22x\)
\(=1-\left(cos2x\right)^2\)
\(=1-\left(2cos^2x-1\right)^2\)
\(=1-\left(2\cdot\dfrac{9}{16}-1\right)^2\)
\(=1-\left(\dfrac{9}{8}-1\right)^2=1-\left(\dfrac{1}{8}\right)^2=\dfrac{63}{64}\)
2:
\(cos2x-sin\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(sin\left(x+\dfrac{\Omega}{3}\right)=cos2x=sin\left(\dfrac{\Omega}{2}-2x\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=\dfrac{\Omega}{2}-2x+k2\Omega\\x+\dfrac{\Omega}{3}=\Omega-\dfrac{\Omega}{2}+2x+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{\Omega}{6}+k2\Omega\\-x=\dfrac{1}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{18}+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{6}\Omega-k2\Omega\end{matrix}\right.\)