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1: ĐKXĐ: 3-cosx>0
=>cosx<3(luôn đúng)
2: ĐKXĐ: 1-sin 3x>=0
=>sin 3x<=1(luôn đúng)
3: ĐKXĐ: sin x<>0 và 2x<>pi/2+kpi
=>x<>kpi và x<>pi/4+kpi/2
4: ĐKXĐ: 2x-1>=0
=>x>=1/2
1. \(sin\left(\dfrac{\pi}{3}-x\right)\ne0\Leftrightarrow\dfrac{\pi}{3}-x\ne k\pi\Leftrightarrow x\ne\dfrac{\pi}{3}-k\pi\)
2. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
3. \(\sqrt{1+sinx}-\sqrt{2}\ge0\Leftrightarrow1+sinx\ge2\Leftrightarrow sinx\ge1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
4. \(\sqrt{2-2cosx}-2\ne0\Leftrightarrow2-2cosx\ne4\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)
5. \(1-\sqrt{1+sin3x}\ne0\Leftrightarrow sin3x\ne0\Leftrightarrow3x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{3}\)
27.
\(\cos\left(\frac{\pi}{6}-2x\right)=\sin x\)
\(\Leftrightarrow\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+2x\right)=\sin x\)
\(\Leftrightarrow\sin\left(\frac{\pi}{3}+2x\right)=\sin x\)
\(\Leftrightarrow\frac{\pi}{3}+2x=\pi-x+k2\pi\Leftrightarrow x=\frac{2}{9}\pi+\frac{2}{3}k\pi\)
\(\frac{\pi}{2}< \frac{2}{9}\pi+\frac{2}{3}k\pi< \pi\Leftrightarrow\frac{5}{18}\pi< \frac{2}{3}k\pi< \frac{7}{9}\pi\)
\(\Leftrightarrow\frac{5}{12}< k< \frac{7}{6}\Rightarrow k=1\)
Vậy phương trình có 1 nghiệm thuộc khoảng \(\left(\frac{\pi}{2};\pi\right)\)
19. \(\sin3x=\sin x\Leftrightarrow3x=\pi-x+k2\pi\Rightarrow x=\frac{\pi}{4}+\frac{1}{2}k\pi\)
33. \(DKXD:\left\{{}\begin{matrix}\cos3x\ne0\\\sin2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne\frac{\pi}{2}+k\pi\\2x\ne\pi+k\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{6}+\frac{1}{3}k\pi\\x\ne\frac{\pi}{2}+\frac{1}{2}k\pi\end{matrix}\right.\)
\(\tan3x.\cot2x=1\Leftrightarrow\tan3x=\frac{1}{\cot2x}=\tan2x\)
\(\Leftrightarrow3x=\pi+2x+k\pi\Leftrightarrow x=\pi+k\pi\) (t/m)
b.
\(\Leftrightarrow\sqrt{2}cos\left(3x+\frac{\pi}{4}\right)=-\sqrt{2}\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow3x+\frac{\pi}{4}=\pi+k2\pi\)
\(\Leftrightarrow x=...\)
c.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=-\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{6}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)