Answer:
Step-by-step explanation:
Given the equation [tex]2\sin^2(x) = 1 \\[/tex], we are to find all the values of x in the interval [0, 2π] that satisfies the equation.
Simplifying the equation:
[tex]2\sin^2(x) = 1 \\\\sin^2x = \frac{1}{2} \\\\[/tex]
Square both sides of the equation;
[tex]\sqrt{sin^2x } = \sqrt{\frac{1}{2} } \\\\sin(x) = \sqrt{0.5}\\ \\sin(x) = 0.7071\\\\x = sin^{-1}0.7071\\\\x \approx 45^0[/tex]
The general formula for sin(x) is x = nπ + (-1)ⁿ∝ where n = 0, 1, 2, 3...
Since ∝ = 45° = π/4.
x = nπ + (-1)ⁿπ/4
when n = 1
x = π + (-1)¹π/4
x = π - π/4
x = 3π/4
Hence the values of x within the given interval are π/4 and 3π/4
