How do I find the exact value of sin2x when secx is equal to the negative square root of three

Answer 1

then cosx = -1/√3

x must be in II or III

in II:
sinx = √2/√3
sin 2x = 2sinxcosx = 2(√2/√3)(-1/√3) = -2√2/3

in III:
sinx = -√2/√3
sin 2x = +2√2/3