giải 1 bài toán nhanh nha mấy bn !!! 1 GP cho câu trả lời đúng :)
giải phương trình : \(sin^3\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sinx\)
Đ/a : \(x=\dfrac{\pi}{4}+k\pi\)
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 phương trình
a) \(sinx=sin\dfrac{\pi}{4}\)
b) \(cos2x=cosx\)
c) \(tan\left(x-\dfrac{\pi}{3}\right)=\sqrt{3}\)
d) \(cot\left(2x+\dfrac{\pi}{6}\right)=cot\dfrac{\pi}{4}\)
a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
b: cos2x=cosx
=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)
c:
ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)
\(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)
=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{2}{3}\Omega+k\Omega\)
d:
ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)
=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)
=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)
\(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)
=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)
=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)
=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)
Giải các pt
a) \(\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2\)
b) \(\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1\)
c) \(2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
c.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)
Phương trình \(\left(\sqrt{3}-1\right)sinx-\left(\sqrt{3}+1\right)cosx+\sqrt{3}-1=0\)có các nghiệm là :
A.\(\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
B.\(\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
C.\(\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{9}+k2\pi\end{matrix}\right.\)
D.\(\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k2\pi\\x=\dfrac{\pi}{12}+k2\pi\end{matrix}\right.\)
Giải một trong 4 đáp án trên hộ em ạ em cảm ơn
Giaỉ các phương trình lượng giác sau:
1. sin(sinx)=0
2. sin(cosx)=0
3. \(\sqrt{3}\sin-\cos x=2cos3x\)
4. \(\sin2x=sin\left(2x-\dfrac{\pi}{2}\right)\)
5. \(4\cos\left(3\pi-2x\right)=\sqrt{2}\)
3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ
1.
\(sin\left(sinx\right)=0\)
\(\Leftrightarrow sinx=k\pi\) (1)
Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)
\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)
Thế vào (1)
\(\Rightarrow sinx=0\Rightarrow x=n\pi\)
2.
\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)
Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)
\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\)
\(\Rightarrow\) Không tồn tại k thỏa mãn
Pt vô nghiệm
giải phương trình
a) \(sin\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
b) \(cos\left(x+\dfrac{\pi}{4}\right)=cos\dfrac{3\pi}{4}\)
c) \(tan2x=tan\left(x+\dfrac{\pi}{3}\right)\)
d) \(cot2x=-\dfrac{\sqrt{3}}{3}\)
a: \(sin\left(x-\dfrac{\Omega}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
=>\(sin\left(x-\dfrac{\Omega}{4}\right)=sin\left(-\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x-\dfrac{\Omega}{4}=-\dfrac{\Omega}{4}+k2\Omega\\x-\dfrac{\Omega}{4}=\Omega+\dfrac{\Omega}{4}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{3}{2}\Omega+k2\Omega\end{matrix}\right.\)
b: \(cos\left(x+\dfrac{\Omega}{4}\right)=cos\left(\dfrac{3}{4}\Omega\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{4}=\dfrac{3}{4}\Omega+k2\Omega\\x+\dfrac{\Omega}{4}=-\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)
c: ĐKXĐ: \(\left\{{}\begin{matrix}2x< >\dfrac{\Omega}{2}+k\Omega\\x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x< >\dfrac{1}{6}\Omega+k\Omega\end{matrix}\right.\)
\(tan2x=tan\left(x+\dfrac{\Omega}{3}\right)\)
=>\(2x=x+\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{\Omega}{3}+k\Omega\)
d: ĐKXĐ: \(2x< >k\Omega\)
=>\(x< >\dfrac{k\Omega}{2}\)
\(cot2x=-\dfrac{\sqrt{3}}{3}\)
=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)
=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)
=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)
giải phương tình sau:
\(\dfrac{1}{\sin x}\)+\(\dfrac{1}{\sin\left(x-\dfrac{3\pi}{2}\right)}\)= 4\(\sin\left(\dfrac{7\pi}{4}-x\right)\)Giải các pt sau
a, \(\dfrac{1}{sinx}+\dfrac{1}{cosx}=4sin\left(x+\dfrac{\pi}{4}\right)\)
b, \(2sin\left(2x-\dfrac{\pi}{6}\right)+4sinx+1=0\)
c, \(cos2x+\sqrt{3}sinx+\sqrt{3}sin2x-cosx=2\)
d, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)