giải phương trình
a) \(sinx=\dfrac{3}{2}\)
b) \(sinx=\dfrac{\sqrt{2}}{2}\)
c) \(sin7x=sin5x\)
d) \(sin5x=sin45\) độ
giải các phương trình sau
a) \(sin5x=sinx\)
b) \(cos\left(x+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)
a: sin 5x=sin x
=>\(\left[{}\begin{matrix}5x=x+k2\Omega\\5x=\Omega-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=k2\Omega\\6x=\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{k\Omega}{2}\\x=\dfrac{\Omega}{6}+\dfrac{k\Omega}{3}\end{matrix}\right.\)
b; \(cos\left(x+\dfrac{\Omega}{3}\right)=-\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=\dfrac{2}{3}\Omega+k2\Omega\\x+\dfrac{\Omega}{3}=-\dfrac{2}{3}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)
giải phương trình
1.\(2sin15x+\sqrt{3}cos5x+sin5x=0\)
2.\(\left(cos2x-\sqrt{3}sin2x\right)-\sqrt{3}sinx-cosx+4=0\)
3.\(cos7x-sin5x=\sqrt{3}\left(cos5x-sin7x\right)\)
4.\(\frac{cosx-2sinx.cosx}{2cos^2x+sinx-1}=\sqrt{3}\)
1.
\(\Leftrightarrow sin5x+\sqrt{3}cos5x=-2sin15x\)
\(\Leftrightarrow\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x=-sin15x\)
\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(-15x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=-15x+k2\pi\\5x+\frac{\pi}{3}=\pi+15x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{60}+\frac{k\pi}{10}\\x=-\frac{\pi}{15}+\frac{k\pi}{5}\end{matrix}\right.\)
2.
\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=2\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)=2\)
Do \(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)\le1\\sin\left(x+\frac{\pi}{6}\right)\le1\end{matrix}\right.\) với mọi x
\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)\le2\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)=1\\sin\left(x+\frac{\pi}{6}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\\x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)
3.
\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)
\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)
sinx - sin3x + sin5x =0
sin2x + sin22x = sin23x
cos3x - cos5x = sinx
sin3x + sin5x + sin7x = 0
sinx + sin2x + sin3x - cosx - cos2x - cos3x = 0
Giải các phương trình sau
a. Cosx+cos2x+cos3x+cos4x=0
b. Sinx+sin3x+sin5x+sin7x=0
\(cosx+cos3x+cos2x+cos4x=0\)
\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)
\(\Leftrightarrow cosx.\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=\pi+k2\pi\end{matrix}\right.\)
\(sinx+sin7x+sin3x+sin5x=0\)
\(\Leftrightarrow2sin4x.cos3x+2sin4x.cosx=0\)
\(\Leftrightarrow sin4x\left(cos3x+cosx\right)=0\)
\(\Leftrightarrow sin4x.cos2x.cosx=0\)
\(\Leftrightarrow sin4x=0\)
\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)
Lý do chỉ cần 1 pt sin4x=0 do sin4x bao hàm cả cosx và cos2x ở trong đó
Chung minh. 1-cos2x/1+cos2x=tan^2x
Bien doi thanh tich
a, A= sina +sinb+sin(a+b)
b, B=cosa +cosb +cos(a+b)+1
c, C= 1 + sina + cosa
d. D = sinx + sin3x +sin5x+sin7x
Chứng minh
a, sinx*sin(pi/3-x)*sin(pi/3+x)=1/4sin3x
b, cosx*cos(pi/3-x)*cos(pi/3+x)=1/4cos3x
c, cos5x*cos3x+sin7x*sinx=cos2x *cos4x
d, sin5x -2sinx(cos2x+cos4x)=sinx
giải phương trình
a) \(sinx=\dfrac{4}{3}\)
b) \(sin2x=-\dfrac{1}{2}\)
c) \(sin\left(x-\dfrac{\pi}{7}\right)\) = \(sin\dfrac{2\pi}{7}\)
d) \(2sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{3}\)
`a)sin x =4/3`
`=>` Ptr vô nghiệm vì `-1 <= sin x <= 1`
`b)sin 2x=-1/2`
`<=>[(2x=-\pi/6+k2\pi),(2x=[7\pi]/6+k2\pi):}`
`<=>[(x=-\pi/12+k\pi),(x=[7\pi]/12+k\pi):}` `(k in ZZ)`
`c)sin(x - \pi/7)=sin` `[2\pi]/7`
`<=>[(x-\pi/7=[2\pi]/7+k2\pi),(x-\pi/7=[5\pi]/7+k2\pi):}`
`<=>[(x=[3\pi]/7+k2\pi),(x=[6\pi]/7+k2\pi):}` `(k in ZZ)`
`d)2sin (x+pi/4)=-\sqrt{3}`
`<=>sin(x+\pi/4)=-\sqrt{3}/2`
`<=>[(x+\pi/4=-\pi/3+k2\pi),(x+\pi/4=[4\pi]/3+k2\pi):}`
`<=>[(x=-[7\pi]/12+k2\pi),(x=[13\pi]/12+k2\pi):}` `(k in ZZ)`
a: sin x=4/3
mà -1<=sinx<=1
nên \(x\in\varnothing\)
b: sin 2x=-1/2
=>2x=-pi/6+k2pi hoặc 2x=7/6pi+k2pi
=>x=-1/12pi+kpi và x=7/12pi+kpi
c: \(sin\left(x-\dfrac{pi}{7}\right)=sin\left(\dfrac{2}{7}pi\right)\)
=>x-pi/7=2/7pi+k2pi hoặc x-pi/7=6/7pi+k2pi
=>x=3/7pi+k2pi và x=pi+k2pi
d: 2*sin(x+pi/4)=-căn 3
=>\(sin\left(x+\dfrac{pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)
=>x+pi/4=-pi/3+k2pi hoặc x-pi/4=4/3pi+k2pi
=>x=-7/12pi+k2pi hoặc x=19/12pi+k2pi
giải phương trình
a) \(sinx=-\dfrac{6}{5}\)
b) \(sin3x=\dfrac{\sqrt{3}}{2}\)
c) \(sin\left(x+\dfrac{\pi}{3}\right)=sin\dfrac{3\pi}{4}\)
d) \(4sin\left(x+\dfrac{5\pi}{6}\right)=5\)
a: sin x=-6/5=-1,2
mà -1<=sin x<=1
nên \(x\in\varnothing\)
b: sin3x=căn 3/2
=>3x=pi/3+k2pi hoặc 3x=2/3pi+k2pi
=>x=pi/9+k2pi/3 hoặc x=2/9pi+k2pi/3
c: \(sin\left(x+\dfrac{pi}{3}\right)=sin\left(\dfrac{3}{4}pi\right)\)
=>x+pi/3=3/4pi+k2pi hoặc x+pi/3=1/4pi+k2pi
=>x=5/12pi+k2pi hoặc x=-1/12pi+k2pi
d: =>sin(x+5/6pi)=5/4
mà sin(x+5/6pi) thuộc [-1;1]
nên \(x\in\varnothing\)
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) \(\tan x+\cot x=2\left(\sin2x+\cos2x\right)\)
Biến đổi tổng thành tích:
A= \(\dfrac{Cosx+Cosy}{Cosx-Cosy}\)
B=\(\dfrac{Sin7x+Sin5x}{Sin7x-Sin5x}\)
\(A=\dfrac{cosx+cosy}{cosx-cosy}=\dfrac{2cos\dfrac{x+y}{2}.cos\dfrac{x-y}{2}}{-2sin\dfrac{x+y}{2}.sin\dfrac{x-y}{2}}=-cot\dfrac{x+y}{2}.cot\dfrac{x-y}{2}\)
\(B=\dfrac{sin7x+sin5x}{sin7x-sin5x}=\dfrac{2sin6x.cosx}{2cos6x.sinx}=tan6x.cotx\)