Question

Tìm tập xác định của hàm số:

y=\(\frac{3sinx+cosx}{cos\left(4x+\frac{2\pi}{5}\right)+cos\left(3x-\frac{\pi}{4}\right)}\)

Tìm giá trị gần đúng nghiệm của các phương trình sau:

cos\(\frac{x}{2}\)= \(\frac{\sqrt{2}}{3}\) trong khoảng (2π, 4π)

Giải các phương trình:

a, cos3x+cos2x-cosx-1=0

b, (2cos-1)(2sinx+cosx)=sin2x-sinx

Answer 1

@Nguyễn Việt Lâm giúp em với ạ

Answer 2

1.

ĐKXĐ: \(cos\left(4x+\frac{2\pi}{5}\right)+cos\left(3x-\frac{\pi}{4}\right)\ne0\)

\(\Leftrightarrow cos\left(4x+\frac{2\pi}{5}\right)\ne cos\left(3x+\frac{3\pi}{4}\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+\frac{2\pi}{5}\ne3x+\frac{3\pi}{4}+k2\pi\\4x+\frac{2\pi}{5}\ne-3x-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{7\pi}{20}+k2\pi\\x\ne-\frac{23\pi}{140}+\frac{k2\pi}{7}\end{matrix}\right.\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=arccos\left(\frac{\sqrt{2}}{3}\right)+k2\pi\\\frac{x}{2}=-arccos\left(\frac{\sqrt{2}}{3}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\left(\frac{\sqrt{2}}{3}\right)+k4\pi\\x=-2arccos\left(\frac{\sqrt{2}}{3}\right)+k4\pi\end{matrix}\right.\)

\(\Rightarrow x=4\pi-2arccos\left(\frac{\sqrt{2}}{3}\right)\approx10.41\left(rad\right)\)

Answer 3

3.

a.

\(\Leftrightarrow\left(cos3x-cosx\right)+\left(cos2x-1\right)=0\)

\(\Leftrightarrow-2sin2x.sinx+1-2sin^2x-1=0\)

\(\Leftrightarrow sin2x.sinx+sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx+sin^2x=0\)

\(\Leftrightarrow sin^2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 4

3b.

\(\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)

\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(2cosx-1\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)