Question

А) sin^2x+sinx-2=0 б)3sin^2x-cosx+1=0

Answer 1

Держи) Если что-то непонятно – обращайся

Answer 2

To solve these equations for x, we can use basic algebraic techniques and trigonometric identities. Let's solve each equation step by step:

a) $\sin^2{x} + \sin{x} - 2 = 0$

To make this equation easier to handle, let's use a substitution. Let $u = \sin{x}$, then the equation becomes:

$u^2 + u - 2 = 0$

Now we can solve this quadratic equation for $u$:

$u^2 + u - 2 = (u + 2)(u - 1) = 0$

Setting each factor to zero and solving for $u$:

1. $u + 2 = 0$ → $u = -2$
2. $u - 1 = 0$ → $u = 1$

Now that we have the possible values for $u$, let's convert them back to $x$:

1. If $u = -2$, then $\sin{x} = -2$. However, this is not possible because the range of the sine function is between -1 and 1.
2. If $u = 1$, then $\sin{x} = 1$. This is possible when $x = \frac{\pi}{2}$ or $x = \frac{5\pi}{2}$ or any other angle that is of the form $\frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

So the solutions for $x$ are $x = \frac{\pi}{2} + 2n\pi$ for any integer $n$.

b) $3\sin^2{x} - \cos{x} + 1 = 0$

Let's try to simplify this equation using trigonometric identities. We can use the identity $\sin^2{x} + \cos^2{x} = 1$ to replace $\cos{x}$ in the equation:

$3\sin^2{x} - (1 - \sin^2{x}) + 1 = 0$

Now simplify further:

$3\sin^2{x} - 1 + \sin^2{x} + 1 = 0$

Combine like terms:

$4\sin^2{x} = 0$

Now we can solve for $\sin{x}$:

$\sin^2{x} = 0$

Taking the square root of both sides:

$\sin{x} = 0$

The solutions for $x$ are $x = 0 + 2n\pi$ or $x = \pi + 2n\pi$ for any integer $n$.