Giải các phương trình sau :
a) \(3\cos^2x-2\sin x+2=0\)
b) \(5\sin^2x+3\cos x+3=0\)
c) \(\sin^6x+\cos^6x=4\cos^22x\)
d) \(-\dfrac{1}{4}+\sin^2x=\cos^4x\)
Cho \(\sin x+\cos x=m\). Tính theo m các biểu thức sau:
1) \(A=\sin^2x+\cos^2x\)
2) \(B=\sin^3x+\cos^3x\)
3) \(C=\sin^4x+\cos^4x\)
4) \(D=\sin^6x+\cos^6x\)
\(sinx+cosx=m\Leftrightarrow\left(sinx+cosx\right)^2=m^2\)
\(\Leftrightarrow1+2sinx.cosx=m^2\Rightarrow sinx.cosx=\dfrac{m^2-1}{2}\)
\(A=sin^2x+cos^2x=1\)
\(B=sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)
\(=m^3-\dfrac{3m\left(m^2-1\right)}{2}=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{3m-m^3}{2}\)
\(C=\left(sin^2+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1-2\left(\dfrac{m^2-1}{2}\right)^2\)
\(D=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3\left(sin^2x+cos^2x\right)\left(sinx.cosx\right)^2\)
\(=1-3\left(\dfrac{m^2-1}{2}\right)^2\)
Chứng minh các biểu thức sau không phụ thuộc vào x:
1, \(A=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
2, \(B=cos^6x+2sin^4x.cos^2x+3sin^2x.cos^4x+sin^4x\)
3, \(C=cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
4, \(D=cos^2x+cos^2\left(x+\dfrac{2\pi}{3}\right)+cos^2\left(\dfrac{2\pi}{3}-x\right)\)
5, \(E=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)-\left(sin^8x+cos^8x\right)\)
6, \(F=cos\left(\pi-x\right)+sin\left(\dfrac{-3\pi}{2}+x\right)-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\dfrac{3\pi}{2}-x\right)\)
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
Chứng minh các biểu thức sau không phụ thuộc vào x:
a) \(A=2\left(cos^6x+sin^6x\right)-3\left(cos^4x+sin^4x\right)\)
b) \(B=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)^2-sin^8x-cos^8x\)
c) \(C=\dfrac{sin^2x}{1+cotgx}+\dfrac{cos^2x}{1+tgx}+sinx.cosx\)
d) \(D=\dfrac{cotg^2a-cos^2x}{cotg^2x}+\dfrac{sinx.cosx}{cotgx}\)
e) \(E=3\left(sin^8x-cos^8x\right)+4\left(cos^6x-2sin^6x\right)+6sin^4x\)
f) \(F=\dfrac{tg^2x}{sin^2x.cos^2x}-\left(1+tg^2x\right)^2\)
Giải các phương trình sau :
a) \(2\sin^2x+\sin x\cos x-3\cos^2x=0\)
b) \(3\sin^2-4\sin x\cos x+5\cos^2x=2\)
c) \(\sin^2x+\sin2x-2\cos^2+5\cos^2x=2\)
d) \(2\cos^2x-3\sqrt{3}\sin2x-4\sin^2x=-4\)
a) Dễ thấy cosx = 0 không thỏa mãn phương trình đã cho nên chiaw phương trình cho cos2x ta được phương trình tương đương 2tan2x + tanx - 3 = 0.
Đặt t = tanx thì phương trình này trở thành
2t2 + t - 3 = 0 ⇔ t ∈ {1 ; }.
Vậy
b) Thay 2 = 2(sin2x + cos2x), phương trình đã cho trở thành
3sin2x - 4sinxcosx + 5cos2x = 2sin2x + 2cos2x
⇔ sin2x - 4sinxcosx + 3cos2x = 0
⇔ tan2x - 4tanx + 3 = 0
⇔
⇔ x = + kπ ; x = arctan3 + kπ, k ∈ Z.
c) Thay sin2x = 2sinxcosx ; = (sin2x + cos2x) vào phương trình đã cho và rút gọn ta được phương trình tương đương
sin2x + 2sinxcosx - cos2x = 0 ⇔ tan2x + 4tanx - 5 = 0 ⇔
⇔ x = + kπ ; x = arctan(-5) + kπ, k ∈ Z.
d) 2cos2x - 3√3sin2x - 4sin2x = -4
⇔ 2cos2x - 3√3sin2x + 4 - 4sin2x = 0
⇔ 6cos2x - 6√3sinxcosx = 0 ⇔ cosx(cosx - √3sinx) = 0
⇔
Giải các Phương trình sau
a) \(sin^4\frac{x}{2}+cos^4\frac{x}{2}=\frac{1}{2}\)
b) \(sin^6x+cos^6x=\frac{7}{16}\)
c) \(sin^6x+cos^6x=cos^22x+\frac{1}{4}\)
d) \(tanx=1-cos2x\)
e) \(tan(2x+\frac\pi3).tan(\frac\pi3-x)=1\)
f) \(tan(x-15^o).cot(x+15^o)=\frac{1}{3}\)
a.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)
\(\Leftrightarrow1-sin^2x=0\)
\(\Leftrightarrow cos^2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
b.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
c.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)
\(\Leftrightarrow3-3sin^22x=4cos^22x\)
\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)
\(\Leftrightarrow3=3+cos^22x\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Giải phương trình:
1.\(cos^3x.cos3x+sin^3x.sin3x=\frac{\sqrt{2}}{4}\)
2.\(cos^34x=cos^3x.cos3x+sin^3x.sin3x\)
3.\(cos^2x-4sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)+2=0\)
4.\(sin^4x+sin^4\left(x+\frac{\pi}{4}\right)=\frac{1}{4}\)
5.\(sin^6x+cos^6x=\frac{5}{6}\left(sin^4x+cos^4x\right)\)
6.\(sin^6x+cos^6x+\frac{1}{2}sinx.cosx=0\)
7.\(\frac{1}{2}\left(sin^4x+cos^4x\right)=sin^2x.cos^2x+sinx.cosx\)
8.\(sin^6x+cos^6x-3cos8x+2=0\)
9.\(sin^4x+cos^4x-2\left(sin^6\frac{x}{2}+cos^6\frac{x}{2}\right)+1=0\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
1.
\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)
\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)
\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)
\(\Leftrightarrow4cos^32x=\sqrt{2}\)
\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
Giải phương trình lượng giác sau
1) 2 cos 2x -\(\sqrt{3}\) = 0
2)\(\sqrt{3}\) tan x + 1 = 0
3) 2 cos2x = 1
4) 6 sin2 x- 13 sin x + 5 = 0
5) 5 cos 2x + 6 cos x + 1 = 0
6 ) 2 cos 2 2x - 3 cos 2x + 1 = 0
7) tan 2 x + ( 1 - \(\sqrt{3}\)) tan x - \(\sqrt{3}\) = 0
8) cos 6x + 2 sin 3x + 3 = 0
9) cos 2x - 4 cos x - 5 = 0
10 ) 3 cos 2 x = 2 sin 2 x + 4 sin x
11) cos 2x + sin2x + 2 cos x + 1 = 0
12) cos 4x + sin 4x + sin 2x = \(\dfrac{5}{2}\)
Chứng minh rằng \(f'\left(x\right)=0;\forall x\in R\) nếu :
a) \(f\left(x\right)=3\left(\sin^4x+\cos^4x\right)-2\left(\sin^6x+\cos^6x\right)\)
b) \(f\left(x\right)=\cos^6x+2\sin^4x.\cos^2x+3\sin^2x\cos^4x+\sin^4x\)
c) \(f\left(x\right)=\cos\left(x-\dfrac{\pi}{3}\right)\cos\left(x+\dfrac{\pi}{4}\right)+\cos\left(x+\dfrac{\pi}{6}\right)\cos\left(x+\dfrac{3\pi}{4}\right)\)
d) \(f\left(x\right)=\cos^2x+\cos^2\left(\dfrac{2\pi}{3}+x\right)+\cos^2\left(\dfrac{2\pi}{3}-x\right)\)
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
Giải các phương trình sau:
1) sin2x + sin23x - 3cos22x = 0
2) sin22x + sin24x = sin26x
3) cos4x - 5sin4x = 1
4) sin24x + sin23x = cos22x +cos2x với x∈(0;π)
5) 4sin3x - 1 = 3 - √3cos3x
6)sin2x = cos22x + cos23x