Tính :
a) \(4\left(\cos24^0+\cos48^0-\cos84^0-\cos12^0\right)\)
b) \(96\sqrt{3}\sin\dfrac{\pi}{48}\cos\dfrac{\pi}{48}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\)
c) \(\tan9^0-\tan63^0+\tan81^0-\tan27^0\)
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a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{3}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k\pi\\2x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+x+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=cos\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+x+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{7\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)
c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)
=>\(cos\left(x-\dfrac{pi}{6}\right)=sin\left(-2x-\dfrac{pi}{3}\right)\)
=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(\dfrac{pi}{2}-x+\dfrac{pi}{6}\right)\)
=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(-x+\dfrac{2}{3}pi\right)\)
=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)
a: cos3x=8
mà -1<=cos3x<=1
nên \(x\in\varnothing\)
b; \(-2\cdot cosx+\sqrt{3}=0\)
=>\(-2\cdot cosx=-\sqrt{3}\)
=>\(cosx=\dfrac{\sqrt{3}}{2}\)
=>x=pi/6+k2pi hoặc x=-pi/6+k2pi
c: cos(3x-pi/6)=0
=>3x-pi/6=pi/2+k2pi
=>3x=2/3pi+k2pi
=>x=2/9pi+k2pi/3
d: cos(x+2/3pi)=cos(pi/5)
=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi
=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi
e: cos^2(3x)=4
=>cos3x=2(loại) hoặc cos3x=-2(loại)
d: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
a: =>sin 4x=cos(x+pi/6)
=>sin 4x=sin(pi/2-x-pi/6)
=>sin 4x=sin(pi/3-x)
=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi
=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3
b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi
=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi
c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
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
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
a: \(cos\left(2x-\dfrac{\Omega}{6}\right)+cos\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)+sin\left(\dfrac{\Omega}{6}-x\right)=0\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=-sin\left(\dfrac{\Omega}{6}-x\right)=sin\left(x-\dfrac{\Omega}{6}\right)\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=cos\left(\dfrac{\Omega}{2}-x+\dfrac{\Omega}{6}\right)\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=cos\left(-x+\dfrac{2}{3}\Omega\right)\)
=>\(\left[{}\begin{matrix}2x-\dfrac{\Omega}{6}=-x+\dfrac{2\Omega}{3}+k2\Omega\\2x-\dfrac{\Omega}{6}=x-\dfrac{2}{3}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{5}{6}\Omega+k2\Omega\\x=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\)
b: \(cos\left(2x+30^0\right)+sin\left(x-30^0\right)=0\)
=>\(cos\left(2x+30^0\right)=-sin\left(x-30^0\right)\)
=>\(cos\left(2x+30^0\right)=sin\left(-x+30^0\right)\)
=>\(cos\left(2x+30^0\right)=cos\left(60^0+x\right)\)
=>\(\left[{}\begin{matrix}2x+30^0=x+60^0+k\cdot360^0\\2x+30^0=-x-60^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\3x=-90^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\x=-30^0+k\cdot120^0\end{matrix}\right.\)
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}cos4x+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}\left(1-2sin^22x\right)+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow...\)
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
\(A=cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\left(-cos\left(\pi-\dfrac{5\pi}{7}\right)\right)=-cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(\Rightarrow A.sin\left(\dfrac{\pi}{7}\right)=-sin\left(\dfrac{\pi}{7}\right).cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=-\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)=-\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=-\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)=\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)
\(\Rightarrow A=\dfrac{1}{8}\)
\(B=\dfrac{\sqrt{3}}{2}.cos48^0.cos24^0.cos12^0\)
\(\Rightarrow B.sin12^0=\dfrac{\sqrt{3}}{2}sin12^0.cos12^0cos24^0.cos48^0\)
\(=\dfrac{\sqrt{3}}{4}sin24^0cos24^0cos48^0=\dfrac{\sqrt{3}}{8}sin48^0.cos48^0\)
\(=\dfrac{\sqrt{3}}{16}sin96^0=\dfrac{\sqrt{3}}{16}cos6^0\)
\(\Rightarrow2B.sin6^0.cos6^0=\dfrac{\sqrt{3}}{16}cos6^0\Rightarrow B=\dfrac{\sqrt{3}}{32.sin6^0}\)
Biểu thức này ko thể rút gọn tiếp được