Question

Giải các phương trình lượng giác sau:

a, sin²x=1

b, \(\frac{sinx-1}{cos2x+1}=0\)

c, sin(3x-\(\frac{\pi}{6}\)) = \(\frac{\sqrt{3}}{2}\)

d, sin(3x-2)=-1

e, sin3x-cos2x=0

f, sin(2x+ \(\frac{\pi}{3}\)) = tan\(\frac{\pi}{3}\)

g, sin(\(3x-\frac{5\pi}{6}\))

Tính giá trị gần đúng của các nghiệm sau:

sin(2x+ \(\frac{\pi}{6}\))= \(\frac{2}{5}\)trong khoảng (\(-\frac{\pi}{3}\); \(\frac{\pi}{6}\))

Answer 1

@Nguyễn Việt Lâm giúp em với ạ

Answer 2

a.

\(\Leftrightarrow1-cos^2x=1\)

\(\Leftrightarrow cos^2x=0\)

\(\Leftrightarrow cosx=0\)

\(\Leftrightarrow x=\frac{\pi}{2}+k\pi\)

b.

ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow sinx-1=0\)

\(\Leftrightarrow sinx=1\)

\(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\) (ktm ĐKXĐ)

Vậy pt vô nghiệm

c.

\(sin\left(3x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

Answer 3

d.

\(\Leftrightarrow3x-2=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{2}{3}-\frac{\pi}{6}+\frac{k2\pi}{3}\)

e.

\(sin3x=cos2x\)

\(\Leftrightarrow sin3x=sin\left(\frac{\pi}{2}-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k2\pi}{5}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

f.

\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=\sqrt{3}\)

Do \(\left\{{}\begin{matrix}sin\left(2x+\frac{\pi}{3}\right)\le1\\\sqrt{3}>1\end{matrix}\right.\) nên pt vô nghiệm

Answer 4

Câu g đề thiếu

Câu 2:

\(sin\left(2x+\frac{\pi}{6}\right)=\frac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=arcsin\left(\frac{2}{5}\right)+k2\pi\\2x+\frac{\pi}{6}=\pi-arcsin\left(\frac{2}{5}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{1}{2}arcsin\left(\frac{2}{5}\right)+k\pi\\x=\frac{5\pi}{12}-\frac{1}{2}arcsin\left(\frac{2}{5}\right)+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\approx-0,056\left(rad\right)\\x\approx1,1\left(rad\right)\end{matrix}\right.\)