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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
a) y=\(sin^4x+cos^4x-3=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-3=-2-\dfrac{1}{2}.sin^22x\)
Có \(0\le sin^22x\le1\)
\(\Leftrightarrow-2\ge y\ge-\dfrac{5}{2}\)
Min xảy ra \(\Leftrightarrow sin^22x=1\Leftrightarrow sin2x=1\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\left(k\in Z\right)\)
\(\Leftrightarrow x=\dfrac{\Pi}{4}+k\Pi\left(k\in Z\right)\)
Max xảy ra \(\Leftrightarrow sin2x=0\Leftrightarrow2x=k\Pi\Leftrightarrow x=\dfrac{k\Pi}{2}\)
b, \(x\in\left[0;\pi\right]\)
=>\(sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\dfrac{\sqrt{2}}{2};1\right]\)
\(\Leftrightarrow2sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\sqrt{2};2\right]\)
\(\Rightarrow\left\{{}\begin{matrix}Miny=-\sqrt{2}\\Maxy=2\end{matrix}\right.\)
Min xảy ra \(\Leftrightarrow x=0\)
Max xảy ra \(\Leftrightarrow x=\dfrac{\pi}{2}\)
1.
\(2sin\left(x+10^o\right)-\sqrt{12}cos\left(x+10^o\right)=3\)
\(\Leftrightarrow\dfrac{1}{2}sin\left(x+10^o\right)-\dfrac{\sqrt{3}}{2}cos\left(x+10^o\right)=\dfrac{3}{4}\)
\(\Leftrightarrow sin\left(x+50^o\right)=\dfrac{3}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+50^o=arcsin\left(\dfrac{3}{4}\right)+k360^o\\x+50^o=180^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-50^o+arcsin\left(\dfrac{3}{4}\right)+k360^o\\x=130^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)
2.
\(\sqrt{3}sin4x-cos4x=\sqrt{3}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin4x-\dfrac{1}{2}cos4x=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\4x-\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)
1: \(P=sin^22x=1-cos^22x\)
\(=1-\left(cos2x\right)^2\)
\(=1-\left(2cos^2x-1\right)^2\)
\(=1-\left(2\cdot\dfrac{9}{16}-1\right)^2\)
\(=1-\left(\dfrac{9}{8}-1\right)^2=1-\left(\dfrac{1}{8}\right)^2=\dfrac{63}{64}\)
2:
\(cos2x-sin\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(sin\left(x+\dfrac{\Omega}{3}\right)=cos2x=sin\left(\dfrac{\Omega}{2}-2x\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=\dfrac{\Omega}{2}-2x+k2\Omega\\x+\dfrac{\Omega}{3}=\Omega-\dfrac{\Omega}{2}+2x+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{\Omega}{6}+k2\Omega\\-x=\dfrac{1}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{18}+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{6}\Omega-k2\Omega\end{matrix}\right.\)
a: =>cos5x=cos(pi/2-3x)
=>5x=pi/2-3x+k2pi hoặc 5x=3x-pi/2+k2pi
=>8x=pi/2+k2pi hoặc 2x=-pi/2+k2pi
=>x=pi/16+kpi/8 hoặc x=-pi/4+kpi
b: sin4x=cos(x+pi/6)
=>sin4x=sin(pi/2-x-pi/6)
=>sin4x=sin(pi/3-x)
=>4x=pi/3-x+k2pi hoặc 4x=pi-pi/3+x+k2pi
=>5x=pi/3+k2pi hoặc 3x=2/3pi+k2pi
=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3
a: -pi/2<a<0
=>sin a<0
=>sin a=-1/căn 5
tan a=-1/2
cot a=-2
b: pi/2<x<pi
=>cosx<0
=>cosx=-4/5
=>tan x=-3/4
cot x=-4/3
c: -pi<x<-pi/2
=>cosx<0 và sin x<0
1+tan^2x=1/cos^2x
=>1/cos^2x=1+16/25=41/25
=>cosx=-5/căn 41
sin x=-6/căn 41
cot x=5/4
g: 180 độ<x<270 độ
=>cosx <0
=>cosx=-4/5
tan x=3/4
cot x=4/3
\(A=sin\left(\dfrac{7}{9}pi\right)+sin\left(\dfrac{pi}{9}\right)-sin\left(\dfrac{5}{9}pi\right)\)
\(=2\cdot sin\left(\dfrac{1}{2}\cdot\dfrac{8}{9}pi\right)\cdot cos\left(\dfrac{1}{2}\cdot\dfrac{6}{9}pi\right)-sin\left(\dfrac{5}{9}pi\right)\)
\(=sin\left(\dfrac{4}{9}pi\right)-sin\left(\dfrac{5}{9}pi\right)\)
\(=2\cdot cos\left(\dfrac{\dfrac{4}{9}pi+\dfrac{5}{9}pi}{2}\right)\cdot sin\left(\dfrac{\dfrac{4}{9}pi-\dfrac{5}{9}pi}{2}\right)\)
=0
a: \(A=sinx\cdot cosx\cdot\left(sin^4x-cos^4x\right)\)
\(=\dfrac{1}{2}\cdot sin2x\cdot\left(sin^2x-cos^2x\right)\)
\(=\dfrac{1}{2}\cdot sin2x\cdot\left(-cos2x\right)\)
\(=-\dfrac{1}{2}\cdot sin2x\cdot cos2x\)
\(=\dfrac{-1}{4}\cdot sin4x=-\dfrac{1}{4}\cdot sin\left(4\cdot\dfrac{pi}{16}\right)=-\dfrac{\sqrt{2}}{8}\)
b: \(B=\left(sin^4x+cos^4x\right)+sinx\cdot cosx\left(sin^2x-cos^2x\right)\)
\(=\left(sin^2x-cos^2x\right)^2+2\cdot\left(sinx\cdot cosx\right)^2+sinx\cdot cosx\left(sin^2x-cos^2x\right)\)
\(=\left(-cos2x\right)^2+2\cdot\left(\dfrac{1}{2}\cdot sin2x\right)^2+\dfrac{1}{2}\cdot sin2x\cdot\left(-cos2x\right)\)
\(=cos^22x+\dfrac{1}{2}\cdot sin^22x-\dfrac{1}{4}\cdot sin4x\)
\(=cos^2\left(2\cdot\dfrac{pi}{48}\right)+\dfrac{1}{2}\cdot sin^2\left(2\cdot\dfrac{pi}{48}\right)-\dfrac{1}{4}\cdot sin\left(4\cdot\dfrac{pi}{48}\right)\)
\(\simeq0.93\)