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a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)
b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)
b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)
c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
ĐKXĐ: ..
\(\frac{sin3x+sinx+sin2x}{cos3x+cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\sqrt{3}\)
\(\Leftrightarrow tan2x=\sqrt{3}\)
\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)
c/
\(\Leftrightarrow\sqrt{3}sin3x-cos3x=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=2x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\pi-2x+\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
e/
\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)
\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)
\(5(sinx+\frac{cos3x+sin3x}{1+2sin2x})=cos2x+3\) (*)
ĐKXĐ: 1 + 2sin2x \(\ne0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x\ne-\frac{\pi}{12}+k\pi\\x\ne\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
Biến đổi riêng biểu thức: \(\frac{cos3x+sin3x}{1+2sin2x}\)
= \(\frac{\left(4cos^3x-3cosx\right)+\left(3sinx-4sin^3x\right)}{1+2sin2x}\)
\(=\frac{4\left(cos^3x-sin^3x\right)+3\left(sinx-cosx\right)}{1+2sin2x}\)
\(=\frac{4\left(cosx-sinx\right)\left(cos^2x+sinx.cosx+sin^2x\right)-3\left(cosx-sinx\right)}{1+2sin2x}\)
= \(\frac{4\left(cosx-sinx\right)\left(1+sinx.cosx\right)-3\left(cosx-sinx\right)}{1+2sin2x}\)
= \(\frac{\left(cosx-sinx\right)\left(4+2sin2x-3\right)}{1+2sin2x}\)
= cosx - sinx
Khi đó:
(*) \(\Leftrightarrow\) 5(sinx + cosx - sinx) - cos2x - 3 = 0
\(\Leftrightarrow5cosx+1-2cos^2x-3=0\)
\(\Leftrightarrow2cos^2x-5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=2\left(-1\le cosx\le1\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\frac{\pi}{3}+k2\pi\)
Với x \(\in\left[0;2\pi\right]\Leftrightarrow\left[{}\begin{matrix}0\le\frac{\pi}{3}+k2\pi\le2\pi\\0\le-\frac{\pi}{3}+k2\pi\le2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\frac{1}{6}\le k\le\frac{5}{6}\\\frac{1}{6}\le k\le\frac{7}{6}\end{matrix}\right.\)
k \(\in Z\Rightarrow\left[{}\begin{matrix}k=0\\k=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}\\x=\frac{5\pi}{3}\end{matrix}\right.\)
Nghiệm của pt (*) thuộc đoạn [0;\(2\pi\)] là:
S = \(\left\{\frac{\pi}{3};\frac{5\pi}{3}\right\}\)