Question

2 cos^2 t+ sin t +1=0

Answer 1

Решение
2 cos^2 t+ sin t +1=0
2*(1 - sin²t) + sint + 1 = 0
2 - 2sin²t + sint + 1 = 0
2sin²t - sint - 3 = 0
sint = y, IyI ≤ 1
2y² - y - 3 = 0
D = 1 + 4*2*3 = 25
y₁ = (1 - 5)/4
y₁ = - 1
y₂ =  (1 + 5)/4
y₂ = 3/2  не удовлетворяет условию  IyI ≤ 1
sint = - 1
t = - π/2 + 2πk, k ∈ Z

Answer 2

To solve the equation 2cos²(t) + sin(t) + 1 = 0, we can rearrange it and apply some trigonometric identities.

Let's start by replacing cos²(t) with 1 - sin²(t) using the identity cos²(t) + sin²(t) = 1:

2(1 - sin²(t)) + sin(t) + 1 = 0

Expanding the equation, we get:

2 - 2sin²(t) + sin(t) + 1 = 0

Rearranging the terms:

-2sin²(t) + sin(t) + 3 = 0

Now, let's substitute sin(t) with a variable, such as x, to simplify the equation:

-2x² + x + 3 = 0

To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, factoring is not straightforward, and completing the square might involve complex numbers, so we'll use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation -2x² + x + 3 = 0, the coefficients are: a = -2 b = 1 c = 3

Applying the quadratic formula:

x = (-(1) ± √((1)² - 4(-2)(3))) / (2(-2)) x = (-1 ± √(1 + 24)) / (-4) x = (-1 ± √25) / (-4) x = (-1 ± 5) / (-4)

This gives us two possible solutions for x:

x₁ = (-1 + 5) / (-4) = 4 / (-4) = -1 x₂ = (-1 - 5) / (-4) = -6 / (-4) = 3/2

Since we set x as sin(t), we substitute x back into the equation:

sin(t) = -1 or sin(t) = 3/2

However, sin(t) can only take values between -1 and 1, so sin(t) = 3/2 is not possible. Therefore, the only solution is sin(t) = -1.

To find the corresponding values of t, we look for angles whose sine is -1. The angle that satisfies this condition is t = -π/2 + 2πk, where k is an integer.

Therefore, the solution to the equation 2cos²(t) + sin(t) + 1 = 0 is:

t = -π/2 + 2πk, where k is an integer.