Giải các phương trình:
1, cosx - \(\sqrt{3}sinx=2cos\left(\frac{\pi}{3}-x\right)\)
2, sin5x +cos5x = \(\sqrt{2}cos13x\)
3, \(\left(3cosx-4sinx-6\right)^2+2=-3\left(3cosx-4sinx-6\right)\)
giải các phương trình sau: ( pt bậc nhất đối với sinx và cosx)
a, \(sinx+cosx=\sqrt{2}sin5x\)
b, \(\sqrt{3}sin2x+sin\left(\frac{\pi}{2}+2x\right)=1\)
c, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx+\sqrt{3}-1=0\)
d, \(3sin^2x+\sqrt{3}sin2x=3\)
e, \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
f,\(8cos2x=\frac{\sqrt{3}}{sinx}+\frac{1}{cosx}\)
g, \(cosx-\sqrt{3}sinx=2cos\left(\frac{\pi}{3}-x\right)\)
h, \(sin5x-cos5x=\sqrt{2}cos13x\)
i, \(\left(3cosx-4sinx+6\right)^2-9cosx+12sinx-16=0\)
\( a){\mathop{\rm sinx}\nolimits} + \cos x = \sqrt 2 \sin 5x\\ \Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .\sin 5x\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 5x\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = 5x + k2\pi \\ x + \dfrac{\pi }{4} = \pi - 5x + k2\pi \end{array} \right.\left( {k \in \mathbb {Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)
\( b)\sqrt 3 \sin 2x + \sin \left( {\dfrac{\pi }{2} + 2x} \right) = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \sin \dfrac{\pi }{2}\cos 2x + \cos \dfrac{\pi }{2}\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + 1.\cos 2x + 0.\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \cos 2x - 1 = 0\\ \Leftrightarrow 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} + 1 - 2{\sin ^2}x - 1 = 0\\ \Leftrightarrow \sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - si{n^2}x = 0\\ \Leftrightarrow {\mathop{\rm sinx}\nolimits} \left( {\sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\mathop{\rm sinx}\nolimits} = 0\\ \sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \sin \left( {\dfrac{\pi }{3} - x} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \dfrac{\pi }{3} - x = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{3} - k\pi \end{array} \right. \)
Nhiều quá @@ Tách ra đi ><
\( d)3{\sin ^2}x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow 2{\sin ^2}x + 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - 3 = 0\\ *sinx = 0 \Rightarrow \text{không là nghiệm phương trình}\\ *sin \ne 0\\ 2 + 2\sqrt 3 \cot x - 3\left( {1 + {{\cot }^2}x} \right) = 0\\ \Leftrightarrow 3{\cot ^2}x - 2\sqrt 3 \cot x + 1 = 0\\ \Leftrightarrow \cot x = \dfrac{{\sqrt 3 }}{3} \Rightarrow x = \dfrac{\pi }{3} + k\pi \)
Giải các phương trình sau:
a) Sinx + \(\sqrt{3}\) Cosx + 2Sin(\(\dfrac{\Pi}{6}\)-x) = \(\sqrt{2}\)
b) 3Cosx - 4Sinx + \(\dfrac{2}{3Cosx-4Sinx-6}\)= 3
c) 8Sinx = \(\dfrac{\sqrt{3}}{Cosx}+\dfrac{1}{Sinx}\)
d) 3Sin3x - \(\sqrt{3}\) Cos9x = 1 + 4Sin33x
e) 5Sin2x - 6Cos2x = 13
f) Cos7x - \(\sqrt{3}\) Sin7x - Sinx = \(\sqrt{3}\) Cos x
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, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)
giải phương trình đối với sin x và cosx
1) 3sinx-4cosx=5
2) \(\sqrt{3}cos2x+sin2x+2sin\left(2x-\frac{\pi}{6}\right)=2\sqrt{2}\)
3) \(cosx+\sqrt{3}sinx+2cos\left(2x+\frac{\pi}{3}\right)=0\)
4) \(2cos\left(2x+\frac{\pi}{6}\right)+4sinxcosx-1=0\)
5) \(\sqrt{3}cos5x-2sin3x.cos2x-sinx=0\)
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.
\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)
\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)
giải phương trình:
1) \(2\sqrt{2}cos^3x\left(x-\frac{\pi}{4}\right)-3cosx-sinx=0\)
2) \(tanx.sin^2x-2sin^2x=3\left(cos2x+sinxcosx\right)\)
3) \(2sin^3x=cosx\)
4) \(6sinx-2cos^3x=\frac{5sin4xcosx}{2cos2x}\)
giải phương trình sau
\(\sqrt{3}sinx+cosx+2cos\left(x-\dfrac{\pi}{3}\right)=2\)
\(\Leftrightarrow2\left(\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx\right)+2cos\left(x-\dfrac{\pi}{3}\right)=2\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
1) \(4cos^24x+2\left(\sqrt{3}+\sqrt{2}\right)cos4x+\sqrt{6}=0\)
2) \(cos4x+2+sin\left(2x+\frac{3\pi}{2}\right)=2cos^2x\)
3) \(sin\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin\left(\frac{\pi}{6}-x\right)=1\)
4) \(2cos\left(4x-\frac{\pi}{3}\right)+4cos2x=-1\)
5) \(cos^22x+cos^23x=sin^2x\)
6) \(sinx+\left(\sqrt{2}-1\right)cosx=1\)
7) \(cos2x-\left(\sqrt{3}+1\right)cosx+\frac{2+\sqrt{3}}{2}=0\)
1.
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)
2.
\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)
\(\Leftrightarrow2cos^22x-cos2x=cos2x\)
\(\Leftrightarrow cos^22x-cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
3.
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
5.
\(cos^22x+\frac{1}{2}+\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos2x\)
\(\Leftrightarrow cos^22x+\frac{1}{2}\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos^22x+cos4x.cos2x=0\)
\(\Leftrightarrow cos2x\left(cos2x+cos4x\right)=0\)
\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
Tìm giá trị lớn nhất và giá trị nhỏ nhất của các hàm số sau:
1,\(y=5-3cosx\)
2,\(y=3cos^2x-2cosx+2\)
3,\(y=cos^2x+2cos2x\)
4,\(y=\sqrt{5-2sin^2x.cos^2x}\)
5,\(y=cos2x-cos\left(2x-\dfrac{\pi}{3}\right)\)
6,\(y=\sqrt{3}sinx-cosx-2\)
7,\(y=2cos^2x-sin2x+5\)
8,\(y=2sin^2x-sin2x+10\)
9,\(y=sin^6x+cos^6x\)