Find all values of in the interval [0, 2π] that satisfy the equation. Enter your answers from smallest to largest. (Enter NONE in any unused answer blanks.) $$ 2\sin^2(x) = 1 $$

Step-by-step explanation:

Given the equation , we are to find all the values of x in the interval  [0, 2π] that satisfies the equation.

Simplifying the equation:

Square both sides of the equation;

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

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: