Question

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.) $$ 2\sin^2(x) = 1 $$

Answer 1

Answer:

π/4 and 3π/4

Step-by-step explanation:

Given the equation $2\sin^2(x) = 1$, we are to find all the values of x in the interval  [0, 2π] that satisfies the equation.

Simplifying the equation:

$2\sin^2(x) = 1 \\\\sin^2x = \frac{1}{2}$

Square both sides of the equation;

$\sqrt{sin^2x } = \sqrt{\frac{1}{2} } \\\\sin(x) = \sqrt{0.5}\\ \\sin(x) = 0.7071\\\\x = sin^{-1}0.7071\\\\x \approx 45^0$

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