sin(4x-10°) = √2/2
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Những câu hỏi liên quan
Chứng minh các biểu thức sau không phụ thuộc vào x: a) A2left(cos^6x+sin^6xright)-3left(cos^4x+sin^4xright)b) B2left(sin^4x+cos^4x+sin^2x.cos^2xright)^2-sin^8x-cos^8xc) Cdfrac{sin^2x}{1+cotgx}+dfrac{cos^2x}{1+tgx}+sinx.cosxd) Ddfrac{cotg^2a-cos^2x}{cotg^2x}+dfrac{sinx.cosx}{cotgx}e) E3left(sin^8x-cos^8xright)+4left(cos^6x-2sin^6xright)+6sin^4xf) Fdfrac{tg^2x}{sin^2x.cos^2x}-left(1+tg^2xright)^2
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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) \(F=\dfrac{tg^2x}{sin^2x.cos^2x}-\left(1+tg^2x\right)^2\)
sqrt{sin^4x+4cos^2x}+sqrt{cos^4x+4sin^2x}
sqrt{left(1-cos^2xright)^2+4cos^2x}+sqrt{left(1-sin^2xright)^2+4sin^2x}
sqrt{cos^4x-2cos^2x+1+4cos^2x}+sqrt{sin^4x-2sin^2x+1+4sin^2x}
sqrt{cos^4x+2cos^2x+1}+sqrt{sin^4x+2sin^2x+1}
sqrt{left(cos^2x+1right)^2}+sqrt{left(sin^2x+1right)^2}
cos^2x+1+sin^2x+13
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\(\sqrt{\sin^4x+4\cos^2x}+\sqrt{\cos^4x+4\sin^2x}\)
=\(\sqrt{\left(1-cos^2x\right)^2+4\cos^2x}+\sqrt{\left(1-sin^2x\right)^2+4\sin^2x}\)
=\(\sqrt{\cos^4x-2\cos^2x+1+4\cos^2x}+\sqrt{\sin^4x-2\sin^2x+1+4\sin^2x}\)
=\(\sqrt{\cos^4x+2\cos^2x+1}+\sqrt{\sin^4x+2\sin^2x+1}\)
=\(\sqrt{\left(cos^2x+1\right)^2}+\sqrt{\left(sin^2x+1\right)^2}\)
=\(cos^2x+1+sin^2x+1=3\)
Giải phương trình:
1.cos^3x.cos3x+sin^3x.sin3xfrac{sqrt{2}}{4}
2.cos^34xcos^3x.cos3x+sin^3x.sin3x
3.cos^2x-4sin^2left(frac{x}{2}-frac{pi}{4}right)+20
4.sin^4x+sin^4left(x+frac{pi}{4}right)frac{1}{4}
5.sin^6x+cos^6xfrac{5}{6}left(sin^4x+cos^4xright)
6.sin^6x+cos^6x+frac{1}{2}sinx.cosx0
7.frac{1}{2}left(sin^4x+cos^4xright)sin^2x.cos^2x+sinx.cosx
8.sin^6x+cos^6x-3cos8x+20
9.sin^4x+cos^4x-2left(sin^6frac{x}{2}+cos^6frac{x}{2}right)+10
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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.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
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6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
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1.
\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)
\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)
\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)
\(\Leftrightarrow4cos^32x=\sqrt{2}\)
\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
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Xem thêm câu trả lời
giá trị biểu thức
A=\(\cos^4x+2\sin^4x\cos^2x+\sin^4x+2\cos^4x\sin^2x+1\)
\(A=\cos^4x+2\sin^2x.\cos^2x\left(\sin^2x+\cos^2x\right)+\sin^4x+1\)
\(=\cos^4x+2\sin^2x.\cos^2x+\sin^4x+1\)
\(=\left(\sin^2x+\cos^2x\right)^2+1=1+1=2\)
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Chứng minh các biểu thức sau không phụ thuộc vào x:
a) \(A=\cos^4x-\sin^4x+2\sin^2x+\tan2x.\cot2x\)
b) \(B=\sqrt{\sin^4x+4\cos^2x}+\sqrt{\cos^4x+4\sin^2x}\)
c) \(C=3\left(\sin^8x-\cos^8x\right)+4\left(\cos^6x-2\sin^6x\right)+6\sin^4x\)
d) \(D=2\left(\sin^4x+\cos^4x+\sin^2x.\cos^2x\right)-\left(\sin^8x+\cos^8x\right)\)
Giảt pt 1,sin(4x-10°) = √2/2 2, cos(2x=7/8 3, tan 2x=tanx 4, cot(x+π/5)=-1 5, cos3x=sin5x
1.
\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))
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2.
Đề không đúng
3.
ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(tan2x=tanx\)
\(\Rightarrow2x=x+k\pi\)
\(\Rightarrow x=k\pi\)
4.
\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)
\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)
\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))
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5.
\(cos3x=sin5x\)
\(\Leftrightarrow sin5x=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) (\(k\in Z\))
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Giải các pt: A, cos(4x + π/3)=✓3/2. ;. B, sin^2x-3sin3x+2=0. ;. C, tan(2x+10°)=√3. ;. D, tanx.cot2x=1
a) \(cos\left(4x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\Rightarrow cos\left(4x+\dfrac{\pi}{3}\right)=cos\dfrac{\pi}{6}\)
\(\Rightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
..... bạn tự tìm x nhé!
b)\(sin^2x-3sin3x+2=0\)\(\Rightarrow sin^2x-3\left(3sinx-4sin^3x\right)+2=0\)
\(\Rightarrow12sin^3x+sin^2x-9sinx+2=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{2}{3}\\sinx=\dfrac{1}{4}\end{matrix}\right.\) \(\Rightarrow\).... bạn tự tìm x nhé!
c)\(tan\left(2x+10^o\right)=\sqrt{3}\Rightarrow tan\left(2x+10^o\right)=tan60^o\)
\(\Rightarrow2x+10^o=60^o+k180^o\)
\(\Rightarrow x=25^o+k90^o\left(k\in Z\right)\)
d) \(tanx\cdot cot2x=1\)
Đk: \(\left\{{}\begin{matrix}cosx\ne0\\sin2x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m\pi\\x\ne m\dfrac{\pi}{2}\end{matrix}\right.\)
Pt: \(\Rightarrow tanx=tan2x\Rightarrow x=2x+k\pi\)
\(\Rightarrow x=k\pi\)
Đối chiếu với đk trên thỏa mãn đk\(\Rightarrow x=k\pi\)
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NN
19 tháng 4 2021 lúc 15:21
Rút gọn các biểu thức sau
1, \(\dfrac{1+\cot x}{1-\cot x}-\dfrac{2+2\cot^2x}{\left(\tan x-1\right)\left(\tan^2x+1\right)}\)
2, \(\sqrt{\sin^4x+6\cos^2x+3\cos^4x}+\sqrt{\cos^4x+6\sin^2x+3\sin^4x}\)
Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý
Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)
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Câu 1 đề vẫn có vấn đề:
\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2\left(1+cot^2x\right)cot^2x}{\left(tanx-1\right)\left(tan^2x+1\right)cot^2x}=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^2x}{tanx-1}\)
\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)
\(=\dfrac{\left(1-cotx\right)\left(1+2cotx+2cot^2x\right)}{1-cotx}=1+2cotx+2cot^2x\)
Có thể coi như ko thể rút gọn tiếp
2.
\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)
\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)
\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)
\(=2\left(cos^2x+sin^2x\right)+2=4\)
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Giải các phương trình :
1) frac{sin^4x+cos^4x}{sin2x}frac{1}{2}left(tan x+cot2xright)
2) frac{1}{sin x}+frac{1}{sinleft(x-frac{3pi}{2}right)}4sinleft(frac{7pi}{4}-xright)
3)2left(cos^42x-sin^42xright)+cos8x-cos4x0
4)frac{sin^4x+cos^4x}{5sin2x}frac{1}{2}cot2x-frac{1}{8sin2x}
5)sin^4x+cos^4x-3sin2x+frac{5}{2}sin^22x0
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Giải các phương trình :
1) \(\frac{\sin^4x+\cos^4x}{\sin2x}=\frac{1}{2}\left(\tan x+\cot2x\right)\)
2) \(\frac{1}{\sin x}+\frac{1}{\sin\left(x-\frac{3\pi}{2}\right)}=4\sin\left(\frac{7\pi}{4}-x\right)\)
3)\(2\left(\cos^42x-\sin^42x\right)+\cos8x-\cos4x=0\)
4)\(\frac{\sin^4x+\cos^4x}{5\sin2x}=\frac{1}{2}\cot2x-\frac{1}{8\sin2x}\)
5)\(\sin^4x+\cos^4x-3\sin2x+\frac{5}{2}\sin^22x=0\)