Question

4sin^2(x)+9cos(x)+5=0

Answer 1

Ответ:

$x = (-1)^{n+1}arcsin\frac{3}{4}+\pi n,\; n \in Z$

Пошаговое объяснение:

$4sin^2x + 9cosx +5 =0$

Воспользуемся основным тригонометрическим тождеством:

$sin^2x+cos^2x = 1 \Rightarrow sin^2x = 1-cos^2x$

Подставим 1 - cos²x вместо sin²x

$4(1-cos^2x) +9cosx+5 = 0\\\\4 - 4cos^2x + 9cosx + 5 =0\\\\-4cos^2x + 9cosx+9 = 0\\\\4cos^2x - 9cosx - 9 = 0$

Сделаем замену t = cos(x), t∈[-1, 1] - область значений косинуса

$4t^2-9t -9=0\\D = 81 + 4\cdot 4 \cdot 9 = 81 +144 = 225\\\\t_1 = \frac{9+15}{8} = \frac{24}{8} = 3 > 1\\\\t_2 = \frac{9-15}{8} = -\frac{6}{8} = -\frac{3}{4}$

Вернём замену:

$sin (x) = -\frac{3}{4}\\\\x = (-1)^{n+1}arcsin\frac{3}{4}+\pi n,\; n \in Z$

Answer 2

To solve the trigonometric equation 4sin^2(x) + 9cos(x) + 5 = 0, we'll use trigonometric identities and algebraic manipulations.

Let's start by making a substitution using the Pythagorean identity: sin^2(x) + cos^2(x) = 1.

Divide the entire equation by 4: sin^2(x) + (9/4)cos(x) + (5/4) = 0

Now, we can make a substitution: Let y = cos(x).

The equation becomes: sin^2(x) + (9/4)y + (5/4) = 0

Now, we can use the identity sin^2(x) = 1 - cos^2(x): 1 - cos^2(x) + (9/4)y + (5/4) = 0

Rearrange the terms: cos^2(x) - (9/4)y - (9/4) = 0

Now, let's use the quadratic formula to solve for y: y = [-b ± √(b^2 - 4ac)] / 2a

Where a = 1, b = -(9/4), and c = -(9/4).

y = [-(9/4) ± √((-(9/4))^2 - 4(1)(-(9/4)))] / 2(1) y = [-(9/4) ± √(81/16 + 36/16)] / 2 y = [-(9/4) ± √(117/16)] / 2

Now, simplify the expression under the square root: y = [-(9/4) ± √(117)/4] / 2

Now, we have two possible values for y:

1. y = (1/4)(-9 + √117)
2. y = (1/4)(-9 - √117)

Next, recall that y = cos(x), so we can find x using the inverse cosine function:

1. cos(x) = (1/4)(-9 + √117) x = arccos((1/4)(-9 + √117)) + 2πn, where n is an integer.

2. cos(x) = (1/4)(-9 - √117) x = arccos((1/4)(-9 - √117)) + 2πn, where n is an integer.

These are the general solutions for the equation 4sin^2(x) + 9cos(x) + 5 = 0. Keep in mind that there might be more solutions depending on the range of x you are interested in.