I will assume your equation says something like
2sin(2πØ) + √3 = 0
if so, then
2sin(2πØ) = -√3
sin(2πØ) = -√3/2
The 'reference angle' is π/3 (I know sin π/3 = sin60° = √3/2)
the sine is negative in quads III and IV
so 2πØ = π+π/3 or 2πØ = 2π - π/3
Ø = 2/3 or 5/6
Solve.
2sin(2ƒÆ) + �ã3 = 0
interval [0, 2ƒÎ)
How do I start this? I feel like maybe Trig Identities, but I'm not really sure.
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