Question

Решите уравнение 2sin^2x-3 sin x+1=0

Answer 1

Замена: t = sin(x)

2t² - 3t + 1 = 0
D = 1, t = 1, t = 0.5. 

Обратная замена:
· sin(x) = 1 ⇒ x =  $\frac{ \pi }{2} + 2 \pi n$, n ∈ Z; 
· sin(x) = 0.5 ⇒ x =  $\frac{ \pi }{6} + 2 \pi n$ и x =  $\frac{5 \pi }{6} + 2 \pi n$, n ∈ Z. 

Ответ: $\frac{ \pi }{2} + 2 \pi n$, $\frac{ \pi }{6} + 2 \pi n$, $\frac{5 \pi }{6} + 2 \pi n$, n ∈ Z. 

Answer 2

To solve the equation 2sin^2x - 3sinx + 1 = 0, we can use the quadratic formula. The equation is in the form of ax^2 + bx + c = 0, where a = 2, b = -3, and c = 1.

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's plug in the values and solve for x:

x = (3 ± √((-3)^2 - 4 * 2 * 1)) / (2 * 2) x = (3 ± √(9 - 8)) / 4 x = (3 ± √1) / 4

Now, we have two possibilities:

1. x = (3 + 1) / 4 x = 4 / 4 x = 1

2. x = (3 - 1) / 4 x = 2 / 4 x = 1/2

So, the solutions to the equation are x = 1 and x = 1/2.