Question

5sinx+2cos^2x=0 пжжжжжжжжжжжжж

Answer 1

2(1 - Sin² x) + 5Sin x - 5 = 0

2 - 2Sin² x + 5Sin x - 5 = 0

-2Sin²x + 5sin x -3 = 0

2Sin² x - 5Sin x +3 = 0

Решаем как квадратное

D = 1

а) Sin x =1

x = π/2 + 2πk, k∈Z

б) Sin x = 6/4

Answer 2

To solve the trigonometric equation $5\sin(x) + 2\cos^2(x) = 0$, we can use trigonometric identities to simplify the equation and find its solutions. Here's the step-by-step solution:

1. Start with the trigonometric identity: $\cos^2(x) + \sin^2(x) = 1$.
2. Solve the identity for $\sin^2(x)$: $\sin^2(x) = 1 - \cos^2(x)$.
3. Substitute the value of $\sin^2(x)$ in the original equation: $5\sin(x) + 2(1 - \cos^2(x)) = 0$.
4. Distribute the $2$ on the right side: $5\sin(x) + 2 - 2\cos^2(x) = 0$.
5. Rearrange the equation: $5\sin(x) - 2\cos^2(x) + 2 = 0$.
6. Substitute $\cos^2(x)$ with $1 - \sin^2(x)$ again: $5\sin(x) - 2(1 - \sin^2(x)) + 2 = 0$.
7. Distribute the $-2$ on the right side: $5\sin(x) - 2 + 2\sin^2(x) + 2 = 0$.
8. Combine like terms: $2\sin^2(x) + 5\sin(x) = 0$.
9. Factor out $\sin(x)$: $\sin(x)(2\sin(x) + 5) = 0$.

Now we have two possible cases:

Case 1: $\sin(x) = 0$ Solving for $x$ in this case, we get $x = 0 + 2\pi n$ or $x = \pi + 2\pi n$, where $n$ is an integer.

Case 2: $2\sin(x) + 5 = 0$ Solving for $\sin(x)$, we get $\sin(x) = -\frac{5}{2}$. However, there are no real solutions for $\sin(x)$ when it's outside the range $[-1, 1]$.

Therefore, the only solutions for the equation $5\sin(x) + 2\cos^2(x) = 0$ are $x = 0 + 2\pi n$ and $x = \pi + 2\pi n$, where $n$ is an integer.