46) Determine solutions for each equation in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian.
Give exact answers where possible.
g) 64 sin2 x − 25 = 0
To solve the equation 64sin^2(x) - 25 = 0, we can start by bringing the constant to the other side of the equation:
64sin^2(x) = 25
Next, we can divide both sides by 64 to isolate sin^2(x):
sin^2(x) = 25/64
Taking the square root of both sides, we get:
sin(x) = ±√(25/64)
Since sin(x) is positive in the first and second quadrants, we can ignore the negative sign:
sin(x) = √(25/64)
Simplifying the square root, we have:
sin(x) = 5/8
To find the value of x, we can take the inverse sine (sin^(-1)) of both sides:
x = sin^(-1)(5/8)
Using a calculator, we find that sin^(-1)(5/8) is approximately 0.8763 radians.
Therefore, the solution to the equation 64sin^2(x) - 25 = 0 in the interval 0 ≤ x ≤ 2π is approximately x = 0.8763 radians.