Question

Sin²x + 2sinX · cosX - 3cos²X =0

Answer 1

$\sin^2x +2\sin x\cos x-3\cos^2x=0$
Разделим обе части уравнения на  $\cos^2x$ , тогда получаем:
$\dfrac{\sin^2x}{\cos^2x} +2\cdot \dfrac{\sin x\cos x}{\cos^2x} -3\cdot \dfrac{\cos^2x}{\cos^2x} =0\\ \\ tg^2x+2tg x-3=0$
Пусть  $tg x=t\,\,\,(t \in \mathbb{R})$ , тогда будем получать такое уравнение:
$t^2+2t-3=0$
Вычислим дискриминант квадратного уравнения:
$D=b^2-4ac=2^2-4\cdot1\cdot(-3)=4+12=16$
$D\ \textgreater \ 0$ , значит уравнение имеет 2 корня.
$t_1= \dfrac{-b+ \sqrt{D} }{2a} = \dfrac{-2+4}{2} =1;\\ \\ t_2= \dfrac{-b- \sqrt{D} }{2a} = \dfrac{-2-4}{2} =-3.$

Обратная замена:
$tg x= 1\\ x=arctg(1)+ \pi n,n \in \mathbb{Z}\\ \boxed{x_1= \dfrac{\pi}{4} + \pi n,n \in \mathbb{Z}}$

$tg x=-3\\ \boxed{x_2=-arctg(3)+\pi n,n \in \mathbb{Z}}$

Answer 2

To solve the equation sin²x + 2sinx · cosx - 3cos²x = 0, we can use some trigonometric identities to simplify it.

Let's start by using the Pythagorean identity sin²x + cos²x = 1:

(1 - cos²x) + 2sinx · cosx - 3cos²x = 0

Expanding the equation:

1 - cos²x + 2sinx · cosx - 3cos²x = 0

Rearranging the terms:

-4cos²x + 2sinx · cosx + 1 = 0

Now, let's focus on the terms involving cosine. We can factor out -2cosx:

-2cosx(2cosx - sinx) + 1 = 0

Now, we have two cases to consider:

Case 1: -2cosx = 0

If -2cosx = 0, then cosx = 0, which implies x = π/2 + nπ, where n is an integer.

Case 2: 2cosx - sinx = 0

To solve this case, we need to use additional methods such as numerical approximation or graphical analysis, as it doesn't have a simple algebraic solution. We can use numerical methods like Newton's method or the bisection method to approximate the value(s) of x that satisfy the equation.

These are the solutions for the equation sin²x + 2sinx · cosx - 3cos²x = 0.