Find all values of in the interval [0, 2π] that

Ответ:

cisneros3

31.01.2022

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

0,0(0 оценок)

waterborn9800

27.12.2022

Creo que $1,280

Step-by-step explanation:

12,000+800=12,800

12,800 x 10% =1,280

0,0(0 оценок)