Question

Cos^2x+3sinx=3..................................​

Answer 1

Решение в фотографии. удачи)

Answer 2

To solve the trigonometric equation $\cos^2x + 3\sin x = 3$, we can use the Pythagorean identity $\sin^2x + \cos^2x = 1$ to rewrite $\cos^2x$ as $1 - \sin^2x$. Then, the equation becomes:

$1 - \sin^2x + 3\sin x = 3$

Now, we can simplify and rearrange the equation:

$-\sin^2x + 3\sin x - 2 = 0$

This is a quadratic equation in terms of $\sin x$. We can solve it by factoring or using the quadratic formula.

Let's use the quadratic formula:

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

In this case, $a = -1$, $b = 3$, and (c = -2. Now, we can plug these values into the formula:

$\sin x = \frac{-3 \pm \sqrt{3^2 - 4(-1)(-2)}}{2(-1)}$

$\sin x = \frac{-3 \pm \sqrt{9 - 8}}{-2}$

$\sin x = \frac{-3 \pm \sqrt{1}}{-2}$

$\sin x = \frac{-3 \pm 1}{-2}$

Now, we have two possible solutions for $\sin x$:

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

However, $\sin x$ is limited to the range [-1, 1], so the second solution is not valid. Therefore, the only solution is:

$\sin x = 1$

To find the corresponding values of $x$, we know that $\sin(\frac{\pi}{2}) = 1$, so:

$x = \frac{\pi}{2} + 2\pi n$

where $n$ is an integer. This equation represents the general solution for the given trigonometric equation.