Question

Giải phương trình

Answer 1

Pt ⇔\(\dfrac{1}{sinx.cosx}=\dfrac{1}{2sin\left(x+\dfrac{\pi}{3}\right)}\)

⇔ sinx.cosx = 2sin\(\left(x+\dfrac{\pi}{3}\right)\)

⇔ \(\dfrac{1}{2}\)sin2x = 2sin\(\left(x+\dfrac{\pi}{3}\right)\)

⇔ sin2x = 4sin\(\left(x+\dfrac{\pi}{3}\right)\)