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

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

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