Given that 0 < or = x < 2pi, solve each of the following equations in radians. sin2x=square root of 3 over 2
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°)
idk anything here
This is confusing
What's trigonometry? What's Kumon.
To solve the equation sin(2x) = √3/2 in the given interval 0 ≤ x < 2π, we need to find the values of x that satisfy this equation.
Let's start by applying the double-angle identity for sine: sin(2x) = 2sin(x)cos(x).
Therefore, the equation sin(2x) = √3/2 can be rewritten as 2sin(x)cos(x) = √3/2.
Now, we know that sin(π/3) = √3/2 and cos(π/3) = 1/2. However, we need to find the values of x in the range 0 ≤ x < 2π.
Using the periodic property of sin and cos, we can find other solutions. Since the values of sin and cos are positive in the first and second quadrants, we need to consider the angles π/3 and 2π/3.
So, our solutions for sin(x) = √3/2 are x = π/3 and x = 2π/3.
However, we need to find the values of x that satisfy sin(2x) = √3/2 in the given range 0 ≤ x < 2π. Therefore, we need to find the values of 2x that correspond to π/3 and 2π/3.
For x = π/3, we have:
2x = 2(π/3) = 2π/3.
For x = 2π/3, we have:
2x = 2(2π/3) = 4π/3.
So, the solutions to the equation sin(2x) = √3/2 in the range 0 ≤ x < 2π are:
x = π/3 and x = 2π/3.