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c/
\(\Leftrightarrow sinx+sin3x+sin2x=cosx+cos3x+cos2x\)
\(\Leftrightarrow2sin2x.cosx+sin2x=2cos2x.cosx+cos2x\)
\(\Leftrightarrow sin2x\left(2cosx+1\right)=cos2x\left(2cosx+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx+1=0\\sin2x=cos2x=sin\left(\frac{\pi}{2}-2x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\2x=\frac{\pi}{2}-2x+k2\pi\\2x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{2\pi}{3}+k2\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\\\end{matrix}\right.\)
b/
\(\Leftrightarrow sin2x+sin6x-\left(cos5x+cosx\right)=0\)
\(\Leftrightarrow2sin4x.cos2x-2cos3x.cos2x=0\)
\(\Leftrightarrow cos2x\left(sin4x-cos3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin4x=cos3x=sin\left(\frac{\pi}{2}-3x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\4x=\frac{\pi}{2}-3x+k2\pi\\4x=3x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow sin2x=sin6x-sin4x\)
\(\Leftrightarrow2sinx.cosx=2cos5x.sinx\)
\(\Leftrightarrow sinx\left(cosx-cos5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos5x=cosx\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\5x=x+k2\pi\\5x=-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\)
a/ Bạn coi lại vế trái đề bài, nhìn không hợp lý
b/ \(\Leftrightarrow\frac{1}{2}sin9x-\frac{1}{2}sinx=\frac{1}{2}sin5x-\frac{1}{2}sinx\)
\(\Leftrightarrow sin9x=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}9x=5x+k2\pi\\9x=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k\pi}{7}\end{matrix}\right.\)
c/ \(\Leftrightarrow sin2x-cos2x=cosx-sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow cos\left(\frac{3\pi}{4}-2x\right)=cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{3\pi}{4}-2x=x+\frac{\pi}{4}+k2\pi\\\frac{3\pi}{4}-2x=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)
a: -1<=sinx<=1
=>5>=-5sinx>=-5
=>11>=-5sinx+6>=1
=>1<=y<=11
\(y_{min}=1\) khi sin x=1
=>x=pi/2+k2pi
\(y_{max}=11\) khi sin x=-1
=>x=-pi/2+k2pi
b: \(-1< =cosx< =1\)
=>\(1>=-cosx>=-1\)
=>\(-3>=-cosx-4>=-5\)
=>\(-3>=y>=-5\)
\(y_{min}=-5\) khi cosx=1
=>x=k2pi
\(y_{max}=-3\) khi cosx=-1
=>x=pi+k2pi
c: \(-1< =cosx< =1\)
=>\(-\sqrt{3}< \sqrt{3}\cdot cosx< =\sqrt{3}\)
=>\(-\sqrt{3}+8< =y< =\sqrt{3}+8\)
\(y_{min}=-\sqrt{3}+8\) khi cosx=-1
=>x=pi+k2pi
\(y_{max}=\sqrt{3}+8\) khi cosx=1
=>x=k2pi
d: \(-1< =cos3x< =1\)
=>\(1>=-cos3x>=-1\)
=>\(16>=y>=14\)
y min=14 khi cos3x=1
=>3x=k2pi
=>x=k2pi/3
y max=16 khi cos3x=-1
=>3x=pi+k2pi
=>x=pi/3+k2pi/3
e: -1<=sin6x<=1
=>-1+2024<=sin6x+2024<=1+2024
=>2023<=y<=2025
y min=2023 khi sin6x=-1
=>6x=-pi/2+k2pi
=>x=-pi/12+kpi/3
y max=2025 khi sin6x=1
=>6x=pi/2+k2pi
=>x=pi/12+kpi/3
\(\Leftrightarrow2cosx+2cos3x=cosx-\sqrt{3}sinx\)
\(\Leftrightarrow\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=-cos3x\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=cos\left(\pi-3x\right)\)
\(\Leftrightarrow...\)
a.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
Chia 2 vế cho cosx:
\(tanx+1=\dfrac{1}{cos^2x}\)
\(\Rightarrow tanx+1=1+tan^2x\)
\(\Rightarrow\left[{}\begin{matrix}tanx=0\\tanx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow2sin2x+2sin^2x=1\)
\(\Leftrightarrow2sin2x=1-2sin^2x\)
\(\Leftrightarrow2sin2x=cos2x\)
\(\Rightarrow tan2x=\dfrac{1}{2}\)
\(\Rightarrow2x=arctan\left(\dfrac{1}{2}\right)+k\pi\)
\(\Rightarrow x=\dfrac{1}{2}arctan\left(\dfrac{1}{2}\right)+\dfrac{k\pi}{2}\)
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
sin 6 x + cos 6 x = 4 cos 2 2 x ⇔ sin 2 x + cos 2 x 3 - 3 sin 2 x . cos 2 x ( sin 2 x + cos 2 x ) = 4 cos 2 2 x
(1-4sin^2 x)sin3x=1/2
1+sin(x/2)sinx-cos(x/2)sin^2 x=2cos^2 (pi/4-x/2)
(sin^3 x.sin3x+cos^3 x.cos3x)/tan(x-pi/6)tan(x+pi/3)=-1/8
2cos^2 (pi/4-3x)-4cos4x-15sin2x=21