Question

4sin2x-5cos2x=4

Answer 1

Lời giải:
PT \(\Leftrightarrow \frac{4}{\sqrt{41}}\sin 2x-\frac{5}{\sqrt{41}}\cos 2x=\frac{4}{\sqrt{41}}\)

\(\Leftrightarrow \cos a\sin 2x-\sin a\cos 2x=\frac{4}{\sqrt{41}}\) (với \(a=\arccos \frac{4}{\sqrt{41}}\), \(a\in (0;\frac{1}{2}\pi)\))

\(\Leftrightarrow \sin (2x-a)=\frac{4}{\sqrt{41}}\)

\(\Leftrightarrow x=\frac{1}{2}(\arcsin \frac{4}{\sqrt{41}}+2k\pi +a)\) hoặc \(x=\frac{1}{2}(\pi -\arcsin \frac{4}{\sqrt{41}}+2k\pi +a)\)