Question

2 sin^2 x -3 sin x -2 = 0​

Answer 1

Замена : sin X =y
2y^2 -3y-2=0
D=\|25=5
y1=4
y2= -1
Sinx=y1
Sinx= 4 уравнение не имеет решений
Sinx=y2
Sinx= - 1
X= - П/2 +2пn, n€Z

Answer 2

The given equation is a quadratic equation in terms of $\sin(x)$:

$2 \sin^2(x) - 3 \sin(x) - 2 = 0$

To solve for $\sin(x)$, we can use the quadratic formula:

$\sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this equation, $a = 2$, $b = -3$, and $c = -2$.

Substituting these values into the quadratic formula:

$\sin(x) = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}$

Simplifying further:

$\sin(x) = \frac{3 \pm \sqrt{9 + 16}}{4}$ $\sin(x) = \frac{3 \pm \sqrt{25}}{4}$ $\sin(x) = \frac{3 \pm 5}{4}$

This gives us two possible solutions for $\sin(x)$:

1. $\sin(x) = \frac{3 + 5}{4} = 2$
2. $\sin(x) = \frac{3 - 5}{4} = -\frac{1}{2}$

However, the value of $\sin(x)$ must be between -1 and 1. Therefore, the second solution, $\sin(x) = -\frac{1}{2}$, is valid.

So, the solution to the equation $2 \sin^2(x) - 3 \sin(x) - 2 = 0$ is $\sin(x) = -\frac{1}{2}$.