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2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)
\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)
Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)
b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)
\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)
\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)
c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)
\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)
\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)
2.
$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$
Vì: $0\leq \sin ^22x\leq 1$
$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$
Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$
3.
$0\leq |\sin x|\leq 1$
$\Rightarrow 3\geq 3-2|\sin x|\geq 1$
Vậy $y_{\min}=1; y_{\max}=3$
c/
\(\Leftrightarrow2cos4x.sin3x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2sin3x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\3x=\frac{\pi}{6}+k2\pi\\3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow6sinx+3cosx+3=sinx-2cosx+3\)
\(\Leftrightarrow sinx+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx=sin4x\)
\(\Leftrightarrow sin\left(\frac{\pi}{3}-x\right)=sin4x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-x+k2\pi\\4x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow2sinx.cosx+4sinx.cos^2x-2sinx=0\)
\(\Leftrightarrow2sinx\left(cosx+2cos^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\2cos^2x+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
1: cot x=-6 nên cosx/sinx=-6
=>cosx=-6*sinx
\(F=\dfrac{sinx-3\cdot cosx}{cosx+2\cdot sinx}=\dfrac{sinx+18\cdot sinx}{-6\cdot sinx+2\cdot sinx}=\dfrac{20}{-4}=-5\)
2: cotx=1
=>cosx/sinx=1
=>cosx=sinx
\(I=\dfrac{sin^3x-4\cdot sin^3x}{sinx+3sinx}=\dfrac{5\cdot sin^3x}{4\cdot sinx}=\dfrac{5}{4}\cdot sin^2x\)
\(1+cot^2x=\dfrac{1}{sin^2x}\)
=>\(\dfrac{1}{sin^2x}=1+1=2\)
=>sin^2=1/2
=>\(I=\dfrac{5}{4}\cdot\dfrac{1}{2}=\dfrac{5}{8}\)
3: cotx=3
=>cosx/sinx=3
=>cosx=3*sinx
1+cot^2x=1/sin^2x
=>\(\dfrac{1}{sin^2x}=1+9=10\)
=>\(sin^2x=\dfrac{1}{10}\)
\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)
\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)
\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)
a,
\(\cos^3x-\sin^3x=\cos x+\sin x\\ < =>\cos^3x-\cos x=\sin^3x-\sin x\\ < =>\cos x\left(\cos^2x-1\right)=\sin x\left(\sin^2x-1\right)\\ < =>\cos x.\left(-\sin^2x\right)=\sin x.\left(-\cos^2x\right)\\ < =>\dfrac{1}{cosx}=\dfrac{1}{sinx}\)
b,
\(2sinx+2\sqrt{3}cosx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\\ < =>2sinx-\dfrac{1}{sinx}=\dfrac{\sqrt{3}}{cosx}-2\sqrt{3}cosx\\ < =>\dfrac{2sin^2x-1}{sinx}=\dfrac{\sqrt{3}.cosx.\left(1-2cos^2x\right)}{cosx}\\ < =>\dfrac{cos2x}{sinx}=\sqrt{3}.cos2x\\ < =>\dfrac{1}{sinx}=\sqrt{3}\)