Question

4синус ²x + 4cинус x + 1 = 0

Answer 1

$4Sin^{2}x+4Sinx+1=0\\\\(2Sinx+1)^{2} =0\\\\2Sinx+1=0\\\\Sinx=-\frac{1}{2}\\\\x=(-1)^{n}arcSin(-\frac{1}{2})+\pi n,n\in Z\\\\x=(-1)^{n+1}arcSin\frac{1}{2}+\pi n,n\in Z\\\\x=(-1)^{n+1}\frac{\pi }{6}+\pi n,n\in Z$

Answer 2

To solve the equation 4sin²(x) + 4sin(x) + 1 = 0, we can use a substitution to simplify the expression. Let's make the substitution y = sin(x):

4sin²(x) + 4sin(x) + 1 = 0

Now, replace sin²(x) with y²:

4y² + 4y + 1 = 0

This is now a quadratic equation in terms of y. To solve it, we can use the quadratic formula:

For a quadratic equation of the form ax² + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b² - 4ac)) / 2a

In our case, a = 4, b = 4, and c = 1:

y = [-(4) ± √(4² - 4(4)(1))] / 2(4) y = [-4 ± √(16 - 16)] / 8 y = [-4 ± √0] / 8 y = [-4 ± 0] / 8

Now, we have two possible values for y:

1. y = (-4 + 0) / 8 = -4 / 8 = -0.5
2. y = (-4 - 0) / 8 = -4 / 8 = -0.5

Since y = sin(x), and the sine function's range is between -1 and 1, there are no real values of x that satisfy y = -0.5. Therefore, the equation 4sin²(x) + 4sin(x) + 1 = 0 has no real solutions.