giúp mk giải bài này vói ạ
giải phương trình
2(sinx + 1)(sin^22x - 3sinx + 1) = sin4x.cosx
\(sin^2x+sin^22x=1\)
Giải giúp mình phương trình này với ???
\(sin^2x+sin^22x=1\)
\(\Leftrightarrow2sin^2x-1+2sin^22x-2=-1\)
\(\Leftrightarrow-cos2x-2cos^22x+1=0\)
\(\Leftrightarrow\left(cos2x+1\right)\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi+k2\pi\\2x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
mọi người giải chi tiết dùm e 3 bài này với
a) \(2cosx-3sinx+2=0\)
b) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\)
c) \(cos\left(2x-15^0\right)+sin\left(2x-15^0\right)=-1\)
a ) \(2cosx-3sinx+2=0\)
\(\Leftrightarrow2cosx-3sinx=-2\)
\(\Leftrightarrow\dfrac{2}{\sqrt{13}}cosx-\dfrac{3}{\sqrt{13}}sinx=-\dfrac{2}{\sqrt{13}}\)
Thấy : \(\left(\dfrac{2}{\sqrt{13}}\right)^2+\left(\dfrac{-3}{\sqrt{13}}\right)^2=1\) nên tồn tại \(\alpha\) t/m :
\(sin\alpha=\dfrac{2}{\sqrt{13}};cos\alpha=\dfrac{-3}{\sqrt{13}}\) . . Khi đó : \(sin\alpha.cosx+cos\alpha.sinx=\dfrac{-2}{\sqrt{13}}\)
\(\Leftrightarrow sin\left(\alpha+x\right)=\dfrac{-2}{\sqrt{13}}\) ( p/t cơ bản )
b ) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\) ( ĐK : \(cosx\ne-1\Leftrightarrow x\ne\left(2k+1\right)\pi\) ; ( k thuộc Z ) )
\(\Leftrightarrow2+2sinx=cosx+1\) \(\Leftrightarrow cosx-2sinx=1\)
Làm giống như a )
c ) \(cos\left(2x-15^o\right)+sin\left(2x-15^o\right)=-1\)
Đặt \(t=2x-15^o\) ; ta có : \(cos t + sin t = -1\)
\(\Leftrightarrow\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)=-1\) \(\Leftrightarrow sin\left(t+\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{4}\right)\)
Xong rồi bn làm tiếp ; chú ý đổi ra độ
bài 1: giải pt
a,\(\frac{cos\left(cos+2sinx\right)+3sinx\left(sinx+\sqrt{2}\right)}{sin2x-1}=1\)
b,\(\frac{sin^22x-2}{sin^22x-4cos^2x}=tan^2x\)
c, \(\frac{1+sin2x+cos2x}{1+cot^2x}=\sqrt{2}sinxsin2x\)
d, \(2tanx+cotx=2sin2x+\frac{1}{sin2x}\)
Giải các phương trình sau:
a/ Cos(x-pi/3)-sin(x-pi/3)=1
b/ Căn 3 sin2x + 2cos^2x = 2 sinx +1
Giúp mk với ạ
a, \(cos\left(x-\dfrac{\pi}{3}\right)-sin\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(x-\dfrac{7\pi}{12}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow x-\dfrac{7\pi}{12}=\pm\dfrac{\pi}{4}+k2\pi\)
...
b, \(\sqrt{3}sin2x+2cos^2x=2sinx+1\)
\(\Leftrightarrow\sqrt{3}sin2x+2cos^2x-1=2sinx\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+k2\pi\\2x+\dfrac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
Các bạn ơi, giúp tớ giải bài tập toán này với ạ
Giải phương trình:
\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
Giải các phương trình sau:
a, \(\dfrac{Sin^2x+Sinx}{Sinx-1}=-2\)
b,\(\dfrac{Cos2x+Sinx}{Sinx-1}+1=0\)
a)Đk:\(sinx\ne1\)
Pt\(\Leftrightarrow sin^2x+sinx=-2\left(sinx-1\right)\)
\(\Leftrightarrow sin^2x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{-3+\sqrt{17}}{2}\left(tm\right)\\sinx=\dfrac{-3-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arcc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\\x=\pi-arc.sin\left(\dfrac{-3+\sqrt{17}}{2}\right)+k2\pi\end{matrix}\right.\)(\(k\in Z\))
b)Đk:\(sinx\ne1\)
Pt \(\Leftrightarrow\dfrac{1-2sin^2x+sinx}{sinx-1}+1=0\)
\(\Leftrightarrow\dfrac{-\left(sinx-1\right)\left(2sinx+1\right)}{sinx-1}+1=0\)
\(\Leftrightarrow-\left(2sinx+1\right)+1=0\)
\(\Leftrightarrow sinx=0\) (tm)
\(\Leftrightarrow x=k\pi,k\in Z\)
Vậy...
giải gấp giúp em chi tiết 3 bài này với ạ
a) sin4x +cos4x = √3
b) 3sinx + √3.cosx = 1
c) √3.cosx + sinx = - √2
a.
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sin4x+\dfrac{\sqrt{2}}{2}cos4x=\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow cos4x.cos\left(\dfrac{\pi}{4}\right)+sin4x.sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{4}=arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\\4x-\dfrac{\pi}{4}=-arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{16}-\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\end{matrix}\right.\)
b.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{6}\)
\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{3}\right)+sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x-\dfrac{\pi}{3}=-arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x=\dfrac{\pi}{3}-arrcos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{6}\right)+sinx.sin\left(\dfrac{\pi}{6}\right)=cos\left(\dfrac{3\pi}{4}\right)\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=cos\left(\dfrac{3\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{3\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
Nghiệm phương trình cos x ( cos x ) + 3 sin x ( sin x + 2 ) sin 2 x - 1 = 1
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