Question

4 cos квадрат x - 8 sin x + 1 = 0

Answer 1

Решение тригонометрических уравнений.

Answer 2

Ответ:

Объяснение:

$4cos^2x-8sinx+1=0\\4(1-sin^2x)-8sinx+1=0\\4-4sin^2x-8sinx+1=0\\-4sin^2x-8sinx+5=0\\]sinx=t\\-4t^2-8t+5=0\\D=64-4*(-4)*5=144\\t1=(8+12):-8=-2.5\\t2=(8-12):-8=0.5\\sinx=0.5\\x1=\frac{\pi }{6}+2\pi k\\x2=\frac{5\pi }{6} +2\pi k$

t1 не удовлетворяет условию так как sinx∈[-1;1]

Answer 3

To solve the equation 4cos^2(x) - 8sin(x) + 1 = 0, we can use trigonometric identities and algebraic manipulations. Here's how you can proceed:

1. Recall the trigonometric identity: sin^2(x) + cos^2(x) = 1. Rewrite the equation in terms of sin(x): 4(1 - sin^2(x)) - 8sin(x) + 1 = 0.

2. Expand the equation: 4 - 4sin^2(x) - 8sin(x) + 1 = 0.

3. Rearrange the terms: -4sin^2(x) - 8sin(x) + 5 = 0.

4. Divide the entire equation by -1 to make the leading coefficient positive: 4sin^2(x) + 8sin(x) - 5 = 0.

5. Let's use a substitution to simplify the equation. Let's define a new variable u as sin(x): 4u^2 + 8u - 5 = 0.

6. Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula: The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by: x = (-b ± √(b^2 - 4ac)) / (2a).

   Applying this to our equation, with a = 4, b = 8, and c = -5, we get: u = (-8 ± √(8^2 - 4 * 4 * -5)) / (2 * 4).

7. Simplify the expression: u = (-8 ± √(64 + 80)) / 8. u = (-8 ± √144) / 8. u = (-8 ± 12) / 8.

8. Solve for u: For the plus sign: u = (-8 + 12) / 8 = 4 / 8 = 1/2.

   For the minus sign: u = (-8 - 12) / 8 = -20 / 8 = -5/2.

9. Recall that u = sin(x). Solve for x: For u = 1/2: sin(x) = 1/2. x = arcsin(1/2). x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.

   For u = -5/2: sin(x) = -5/2. Since the range of sine function is [-1, 1], there are no solutions for this case.

Therefore, the solutions for the equation 4cos^2(x) - 8sin(x) + 1 = 0 are: x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.