Giải phương trình: \(\sin^3x\cos3x+\cos^3x\sin3x=-\frac{3}{4}\)
\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=3\)
\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=cos^2x-\frac{cos3x}{cosx}+sin^2x+\frac{sin3x}{sinx}\)
\(=1+\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=1+\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}\)
\(=1+\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=1+\frac{2sin2x}{sin2x}=1+2=3\)
\(\cos3x\cos^3x-\sin3x\sin^3x=\frac{2+3\sqrt{2}}{8}\)
\(\frac{cos^2x-cos3x}{cosx}-\frac{sin^3x+sin3x}{sinx}=3\)
Bạn coi lại đề
1 tử số là \(cos^2x\), 1 tử số là \(sin^3x\) hoàn toàn ko hợp lý
Tìm nghiệm của các phương trinh:
1,\(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
2,\(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}\left(1+cot2xcotx\right)=0\)
3,\(cos^4x+sin^4x+cos\left(x-\dfrac{\pi}{4}\right)sin\left(3x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
4,\(cos5x+cos2x+2sin3xsin2x=0\) trên \(\left[0;2\pi\right]\)
5,\(\dfrac{cos\left(cosx+2sinx\right)+3sinx\left(sinx+\sqrt{2}\right)}{sin2x-1}=1\)
6,\(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
7,\(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
5, \(\dfrac{cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx}{sin2x-1}=1\)
⇒ \(cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx=sin2x-1\)
⇒ cos2x + 3sin2x + 3\(\sqrt{2}\)sin2x + 1 = 0
⇔ 2 + 2sin2x + 3\(\sqrt{2}\)sin2x = 0
⇔ 2 + 1 - cos2x + 3\(\sqrt{2}\) sin2x = 0
⇔ \(3\sqrt{2}sin2x-cos2x=-1\)
Còn lại tự giải
7, \(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
⇔ \(2cos2x.cos\dfrac{\pi}{4}+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
⇔ \(\sqrt{2}cos2x+4sinx=2+\sqrt{2}-\sqrt{2}sinx\)
Dùng công thức : cos2x = 1 - 2sin2x đưa về phương trình bậc 2 ẩn sinx
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 :
1. sin( x+\(\pi\)/4)=2/3
2.cos2x-5sinx-3=0
3.cos3x=sin2x
4.cos3x=-\(\sqrt{ }\)3 với -\(\pi\)/2<x<0
5.4sin\(^4\)x + 12cos\(^2\)x=7
6.cot(x-1)=(cos2x)/(1+tanx) + sin\(^2\)x - 1/2sin2x
7.sin\(^2\)3x-cos\(^2\)4x=sin\(^2\)5x-cos\(^2\)6x
1,Giải phương trình:
a,\(cos^3x+sin^3x=cos2x\)
b,\(cos^3x+sin^3x=2sin2x+sinx+cosx\)
c,\(2cos^3x=sin3x\)
d,\(cos^2x-\sqrt{3}sin2x=1+sin^2x\)
e,\(cos^3x+sin^3x=2\left(cos^5x+sin^5x\right)\)
a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)
b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x
⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x
⇔ 4sin2x + (sinx + cosx) . sin2x = 0
⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)
⇔ sin2x = 0
c, 2cos3x = sin3x
⇔ 2cos3x = 3sinx - 4sin3x
⇔ 4sin3x + 2cos3x - 3sinx(sin2x + cos2x) = 0
⇔ sin3x + 2cos3x - 3sinx.cos2x = 0
Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình
Xét cosx ≠ 0 chia cả 2 vế cho cos3x ta được :
tan3x + 2 - 3tanx = 0
⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
d, cos2x - \(\sqrt{3}sin2x\) = 1 + sin2x
⇔ cos2x - sin2x - \(\sqrt{3}sin2x\) = 1
⇔ cos2x - \(\sqrt{3}sin2x\) = 1
⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)
⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)
e, cos3x + sin3x = 2cos5x + 2sin5x
⇔ cos3x (1 - 2cos2x) + sin3x (1 - 2sin2x) = 0
⇔ cos3x . (- cos2x) + sin3x . cos2x = 0
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
Chứng minh đẳng thức sau
\(\dfrac{cos^3x-cos3x}{cosx} + \dfrac{sin^3x+sin3x}{sinx} = 3\)
\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=cos^2x-\frac{cos3x}{cosx}+sin^2x+\frac{sin3x}{sinx}\)
\(=1+\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=1+\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=1+\frac{2sin2x}{sin2x}=3\)
Giải các phương trình sau
a) \(sin^6x+cos^6x=cos2x+\dfrac{1}{16}\)
b) \(sin^4\dfrac{x}{2}+cos^4\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
c) \(cos5xcosx=cos4xcos2x+4-3sin^2x\)
d) \(2cosxcos2x=1+cos2x+cos3x\)
e) \(sin3x+cos2x=2\left(sin2xcosx-1\right)\)
a.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)
\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)
b.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
c.
\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)
\(\Leftrightarrow cos^22x-2cos2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)