Question

Sin^2 x + 6cos^2 x - sinx = 0​

Answer 1

Решение во вложение. Удачи

Answer 2

To solve the equation sin^2(x) + 6cos^2(x) - sin(x) = 0, we can first try to simplify it using trigonometric identities.

Recall the Pythagorean identity: sin^2(x) + cos^2(x) = 1. We can rewrite cos^2(x) as 1 - sin^2(x) using this identity.

Now, the equation becomes:

sin^2(x) + 6(1 - sin^2(x)) - sin(x) = 0.

Next, distribute the 6:

sin^2(x) + 6 - 6sin^2(x) - sin(x) = 0.

Now, combine the sin^2(x) terms:

5sin^2(x) - sin(x) + 6 = 0.

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

The general form of a quadratic equation is ax^2 + bx + c = 0.

In our equation, a = 5, b = -1, and c = 6.

The quadratic formula states that:

x = (-b ± √(b^2 - 4ac)) / 2a.

Substitute the values:

sin(x) = [-( -1) ± √((-1)^2 - 456)] / 2*5 sin(x) = (1 ± √(1 - 120)) / 10 sin(x) = (1 ± √(-119)) / 10.

Since the square root of a negative number is not a real number, the solutions for sin(x) are not real numbers either. Therefore, there are no real solutions to the original equation sin^2(x) + 6cos^2(x) - sin(x) = 0.