Đề thiếu bạn nhé!
\(5(sinx+\frac{cos3x+sin3x}{1+2sin2x})=cos2x+3\) (*)
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 = \(\left\{\frac{\pi}{3};\frac{5\pi}{3}\right\}\)
