Tìm ∫ ( cos 6 x - cos 4 x ) d x l à :
A. - 1 6 sin 6 x + 1 4 sin 4 x + C
B. 6 sin 6 x - 5 sin 4 x + C
C. 1 6 sin 6 x - 1 4 sin 4 x + C
D. - 6 sin 6 x + sin 4 x + C
Những câu hỏi liên quan
Giải các pt sau:
a) \(\cos^2x-\cos x=0\)
b) \(2\sin2x\) + \(\sqrt{2}\sin4x=0\)
c) \(8\cos^2x+2\sin x-7=0\)
d) \(4\cos^4x+\cos^2x-3=0\)
e) \(\sqrt{3}\tan x-6\cot x+\left(2\sqrt{3}-3\right)=0\)
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
Đúng 1
Bình luận (0)
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)
Vậy...
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
Vậy...
c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
Vậy...
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow cos2x=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
Vậy...
e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))
\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
Vậy...
Đúng 1
Bình luận (0)
c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow-8sin^2x+2sinx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\)
Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
Với \(sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0\)
\(\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)\)
\(\Leftrightarrow cos^2x=\dfrac{3}{4}\)
\(\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}\)
Với \(cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)
Với \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
\(\text{2}\sin^3x-3\sin^{\text{2}}x\cos x+4\sin x\cos^{\text{2}}x-\sin x\cos x+6\cos^3x=7\)
Tính:F=Cos(π/4+α) x cos(π/4-α)
G=Sin(π/3+α) x cos(π/3-α)
H=cos(π/2-α) x sin(π/2+α)
I=sin(π/4+α) - cos(π/4-α)
K=cos(π/6-x) - sin(π/3+x)
Tìm giới hạn : \(L=\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{\cos x}-\sqrt[5]{\cos x}}{\sin^2x}\)
Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :
\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)
\(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+.....+y+1\right)}=-\frac{1}{40}\)
Đúng 0
Bình luận (0)
Chứng minh biểu thức sau không phụ thuộc vào $x$.
$P=\sqrt{\sin ^{4} x+6 \cos ^{2} x+3 \cos ^{4} x}+\sqrt{\cos ^{4} x+6 \sin ^{2} x+3 \sin ^{4} x}$.
Tìm hiệu của số tròn chục lớn nhất có 2 chữ số
17 tháng 5 2021 lúc 19:59
Tìm số đo góc nhọn x:
a) \(4\sin x-1=1\)
b) \(2\sqrt{3}-3\tan x=\sqrt{3}\)
c) \(7\sin-3\cos\left(90^o-x\right)=2,5\)
d) \(\left(2\sin-\sqrt{2}\right)\left(4\cos-5\right)=0\)
e) \(\dfrac{1}{\cos^2x}-\tan x=1\)
f) \(\cos^2x-3\sin^2x=0,19\)
a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow x=30^o\)
b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)
\(\Leftrightarrow x=30^o\)
c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)
d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)
Đúng 2
Bình luận (0)
Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(
e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)
f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)
Đúng 2
Bình luận (0)
Chứng minh rằng fleft(xright)0;forall xin R nếu :
a) fleft(xright)3left(sin^4x+cos^4xright)-2left(sin^6x+cos^6xright)
b) fleft(xright)cos^6x+2sin^4x.cos^2x+3sin^2xcos^4x+sin^4x
c) fleft(xright)cosleft(x-dfrac{pi}{3}right)cosleft(x+dfrac{pi}{4}right)+cosleft(x+dfrac{pi}{6}right)cosleft(x+dfrac{3pi}{4}right)
d) fleft(xright)cos^2x+cos^2left(dfrac{2pi}{3}+xright)+cos^2left(dfrac{2pi}{3}-xright)
Đọc tiếp
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
1. CM:
\(\dfrac{1}{2}\le\dfrac{\sin x+2\cos x+3}{2\sin x\cos x+3}\le2\)
2. Giải PT:
a) \(\dfrac{1}{\cos x}=4\sin x+6\cos x\)
b) \(\sin^3\left(x-\dfrac{\pi}{4}\right)=\sqrt{2}\sin x\)
c) \(\dfrac{1}{\cos x}+\dfrac{1}{\sin2x}=\dfrac{2}{\sin4x}\)
1.
Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng
2.a
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)
\(\Leftrightarrow1+tan^2x=4tanx+6\)
\(\Leftrightarrow tan^2x-4tanx-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)
Đúng 4
Bình luận (0)
2b.
Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)
\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint+cost\)
\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)
\(\Leftrightarrow sint.cos^2t+cost=0\)
\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
Đúng 4
Bình luận (0)
2c.
ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)
\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)
\(\Leftrightarrow1-2sin^2x-sinx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)
Đúng 3
Bình luận (0)
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}
Đọc tiếp
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 = \(\dfrac{5}{2}\)
Tìm GTLN - GTNN
1 . \(y=S\times\left(1-\frac{S^2-1}{2}\right)\)
2. \(y=\sin^4x+\cos^4x\)
3.\(y=\sin^6+\cos^6\)
4.\(y=\frac{\cos x+2\sin x+3}{2\cos x-\sin x+4}\)