Question

4sin^2 x + 5sin x < 0

Answer 1

4sin²x + 5sinx < 0

sinx(4sinx + 5) < 0

Найдем нули:

sinx = 0

4sinx + 5 = 0

sinx = -5/4 -- не удовл., т.к. -1 ≤ sinx ≤ 1

=> sinx < 0

x ∈ (π + 2πn; 2π + 2πn)

Answer 2

To solve the inequality $4\sin^2 x + 5\sin x < 0$, we need to find the values of $x$ that satisfy the inequality. Here's how we can do that:

Step 1: Factor the expression if possible. The inequality $4\sin^2 x + 5\sin x < 0$ doesn't factor easily, so we'll use an alternative approach.

Step 2: Rewrite the inequality in terms of a single trigonometric function. Notice that the inequality involves only one trigonometric function, $\sin x$. We can simplify it by factoring out $\sin x$:

$\sin x(4\sin x + 5) < 0$

Step 3: Find critical points. The inequality is satisfied when either $\sin x < 0$ or $(4\sin x + 5) < 0$. Let's find the critical points for both conditions:

Condition 1: $\sin x < 0$ $\sin x$ is negative in the third and fourth quadrants.

Condition 2: $4\sin x + 5 < 0$ Subtract 5 from both sides: $4\sin x < -5$ Divide by 4 (remembering to reverse the inequality since we are dividing by a negative number): $\sin x < -\frac{5}{4}$

Step 4: Find the solution in the given range. The range for $x$ is not specified in the question, so we'll consider the full range of $x$.

For $\sin x < 0$, $x$ should be in the third and fourth quadrants.

For $\sin x < -\frac{5}{4}$, $x$ should also be in the third and fourth quadrants.

Step 5: Final solution The solution to the inequality $4\sin^2 x + 5\sin x < 0$ is the intersection of the solutions for $\sin x < 0$ and $\sin x < -\frac{5}{4}$. Therefore, the solution is all values of $x$ in the third and fourth quadrants.

In interval notation, the solution is: $x \in \left(-\frac{\pi}{2}, 0\right) \cup \left(\pi, \frac{3\pi}{2}\right)$

In degrees, the solution is: $x \in \left(-90^\circ, 0^\circ\right) \cup \left(180^\circ, 270^\circ\right)$