Question

giải pt

a) \(sin2x-\sqrt{3}cos2x=0\)

b) \(cos\left(90^o-x\right)+sin2x=0\)

c) \(sinx+\sqrt{3}cosx=0\)

Answer 1

\(sin2x=\sqrt{3}cos2x\)

Nhận thấy cos2x=0 ko phải nghiệm, pt tương đương:

\(\frac{sin2x}{cos2x}=\sqrt{3}\Leftrightarrow tan2x=\sqrt{3}\)

\(\Rightarrow2x=\frac{\pi}{3}+k\pi\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

b/

\(cos\left(90^0-x\right)=-sin2x=cos\left(2x+90^0\right)\)

\(\Rightarrow\left[{}\begin{matrix}90^0-x=2x+90^0+k360^0\\90^0-x=-2x-90^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k120^0\\x=-180^0+k360^0\end{matrix}\right.\)

c/ Giống câu a

\(\Leftrightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)