Given that 0 < or = x < 2pi, solve each of the following equations in radians. sin2x=square root of 3 over 2
5 answers
Do you go to Kumon? I work at a Kumon. You should ask your instructor for a answer guide, which takes you step by step on the higher levels for when you get stuck. You should be able to get one, since you are on the upper levels.
sin 2x = √3/2 , and the sine is + in I and II
so
2x = 60° (π/3) or 2x = 120° (2π/3)
so
x = π/6 or x = π/3
the period of sin 2x is π, so adding π to our two answers would yield two more in the given domain.
x = π/6 + π = 7π/6 or x = π/3 + π = 4π/3
x = π/6 , π/3 , 7π/6, and 4π/3
(30°, 60°, 210°, 240°)
so
2x = 60° (π/3) or 2x = 120° (2π/3)
so
x = π/6 or x = π/3
the period of sin 2x is π, so adding π to our two answers would yield two more in the given domain.
x = π/6 + π = 7π/6 or x = π/3 + π = 4π/3
x = π/6 , π/3 , 7π/6, and 4π/3
(30°, 60°, 210°, 240°)
idk anything here
This is confusing
What's trigonometry? What's Kumon.
2 answers