Question

2 sin^x + 3 sin x =2

Answer 1

Ответ:

0

Пошаговое объяснение:

мб 2sin2x + 3sinx – 2 = 0

sinx = t

2t2 + 3t – 2 = 0

t=–2, t=1/2

–2 не подходит так как |t|<=1

sinx = 1/2

x=π/6 + 2πn, n ∈ Z

x=5π/6 + 2πn, n ∈ Z

Answer 2

To solve the equation $2\sin^2(x) + 3\sin(x) = 2$, we can first rewrite it in terms of a quadratic equation. Let $u = \sin(x)$, then the equation becomes:

$2u^2 + 3u - 2 = 0$

Now, we can solve this quadratic equation for $u$. You can use the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 2$, $b = 3$, and $c = -2$. Plugging these values into the formula:

$u = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}$

Simplify the expression under the square root:

$u = \frac{-3 \pm \sqrt{9 + 16}}{4}$

$u = \frac{-3 \pm \sqrt{25}}{4}$

$u = \frac{-3 \pm 5}{4}$

Now, we have two possible solutions for $u$:

1. $u_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$
2. $u_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$

Remember that $u = \sin(x)$, so we need to find the corresponding values of $x$ for these values of $u$.

1. For $u_1 = \frac{1}{2}$, you can use the inverse sine function to find $x$:

$\sin(x) = \frac{1}{2}$

This occurs when $x$ is either $\frac{\pi}{6} + 2\pi n$ or $\frac{5\pi}{6} + 2\pi n$, where $n$ is an integer.

2. For $u_2 = -2$, there is no solution since the sine function's range is between -1 and 1, and it can never equal -2.

So, the solutions for the equation $2\sin^2(x) + 3\sin(x) = 2$ are:

$x = \frac{\pi}{6} + 2\pi n$ or $x = \frac{5\pi}{6} + 2\pi n$, where $n$ is an integer.