Question

Sin^2 (2x) +2 sin (x) - 3=0

Answer 1

$\sin^{2} 2x + 2\sin x - 3 = 0\\4\sin^{2} x \cos^{2} x + 2\sin x - 3 = 0\\4\sin^{2} x (1 - \sin^{2}x) + 2\sin x - 3 = 0\\4\sin^{2} x -4\sin^{4} x + 2\sin x - 3 = 0$

Замена: $\sin x = t, \ t \in [-1; \ 1]$

$4t^{2} - 4t^{4} + 2t - 3 = 0$

Решим уравнение графически, для этого преобразуем его.

$2t - 3 = 4t^{4} - 4t^{2}\\2t - 3 + 1 = 4t^{4} - 4t^{2} + 1\\2t - 2 = (2t^{2} - 1)^{2}\\\sqrt{2t - 2} = \sqrt{(2t^{2} - 1)^{2}}\\\sqrt{2t - 2} =|2t^{2} - 1|$

Рассмотрим две функции: $y = \sqrt{2t -2}$ и $y = |2t^{2} - 1|$. Изобразим их на координатной плоскости (см. вложение). Видим, что нет точек пересечения. Делаем вывод: заданное уравнение не имеет действительных корней.

Ответ: нет действительных корней.

Answer 2

To solve the trigonometric equation sin^2(2x) + 2sin(x) - 3 = 0, we can use substitution and quadratic factorization. Let's start by making a substitution:

Let y = sin(x).

Now, the equation becomes a quadratic equation in terms of y:

sin^2(2x) + 2sin(x) - 3 = 0 (sin(2x))^2 + 2sin(x) - 3 = 0 (2sin(x)cos(x))^2 + 2sin(x) - 3 = 0 (2y * sqrt(1 - y^2))^2 + 2y - 3 = 0 4y^2(1 - y^2) + 2y - 3 = 0

Now, let's solve the quadratic equation 4y^2(1 - y^2) + 2y - 3 = 0:

4y^2 - 4y^4 + 2y - 3 = 0

This equation is a quartic equation, but we can attempt to factor it if possible. Unfortunately, it does not have a simple factorization. Therefore, to solve for y, we'll use numerical methods like graphing or numerical approximation techniques.

For example, let's plot the equation y = sin(x) and y = (3 - 4y^2)^0.25:

By plotting these two equations, the x-values where the graphs intersect will give us the values of y (sin(x)) that satisfy the original equation.

Please note that this method gives an approximate solution. If you need more precise solutions, numerical methods can be employed.