Question

4sin^2 x- cosx + 1=0

Answer 1

Ответ:

x=2kπ? k ЄZ  

или же

x=k⋅360°,kЄZ

Answer 2

Ответ:

Объяснение:

X=k×360°, k<z

Answer 3

To solve the equation 4sin^2(x) - cos(x) + 1 = 0, we can use trigonometric identities to simplify it. Let's start by making a substitution:

Let y = sin(x)

Now, we can rewrite the equation in terms of y:

4y^2 - cos(x) + 1 = 0

Next, we need to express cos(x) in terms of y. We can use the trigonometric identity:

cos^2(x) + sin^2(x) = 1

From this identity, we can solve for cos(x):

cos^2(x) = 1 - sin^2(x)

cos(x) = ±√(1 - sin^2(x))

Now, substitute this into the equation:

4y^2 - ±√(1 - y^2) + 1 = 0

To solve for y, you can use the quadratic formula:

y = [-b ± √(b^2 - 4ac)] / (2a)

In this case, a = 4, b = ±1, and c = 1. Plug in these values:

y = [±1 ± √((-1)^2 - 4(4)(1))] / (2(4))

y = [±1 ± √(1 - 16)] / 8

y = [±1 ± √(-15)] / 8

Since the square root of a negative number isn't a real number, this equation has no real solutions. Therefore, the original equation 4sin^2(x) - cos(x) + 1 = 0 has no real solutions for x.