Question

giải pt sau a) sin2x.tan(x-pi/4)=0

b)cot9x=-tan(pi/9+9x)

c)cos(50*+4x)+sin3x=0

Answer 1

a/ ĐKXĐ: \(x\ne\frac{3\pi}{4}+k\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\tan\left(x-\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x-\frac{\pi}{4}=k\pi\end{matrix}\right.\)

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

b/

\(\Leftrightarrow sin3x=-cos\left(4x+50^0\right)\)

\(\Leftrightarrow sin3x=sin\left(4x-40^0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-40^0=3x+k360^0\\4x-40^0=180^0-3x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40^0+k360^0\\x=\frac{220^0}{7}+\frac{k360^0}{7}\end{matrix}\right.\)