Find a numerical value of one trigonometric function of x for cos²x + 2sinx - 2 = 0

Solution:

Given cos²x + 2sinx - 2 = 0

We know trignometric identity: sin²x + cos²x = 1

So, cos²x = 1 - sin²x

⇒ (1 - sin²x) + 2sinx - 2 = 0

⇒ -sin²x + 2sinx - 1 = 0

⇒ sin²x - 2sinx + 1 = 0

This is similar to the algebraic identity: (a² - 2ab + b²) = (a - b)²

⇒ (sinx - 1)(sinx - 1) = 0

⇒ (sinx - 1)² = 0

⇒ sinx = 1

⇒ x = sin^-1(1)

⇒ x = π/2 + 2πn

Find a numerical value of one trigonometric function of x for cos²x + 2sinx - 2 = 0

Summary:

A numerical value of one trigonometric function of x for cos²x + 2sinx - 2 = 0 is x = π/2 + 2πn.