Question

(2cos^2x - 5cosx + 2)*корень из sinx =0

Answer 1

Решаем с учётом ОДЗ.

Answer 2

To solve the equation (2cos^2x - 5cosx + 2)*корень из sinx = 0, we need to find the values of x that satisfy the equation. Let's break down the equation and solve it step by step.

The equation can be written as:

(2cos^2x - 5cosx + 2)√(sinx) = 0

To solve this equation, we can set each factor to zero and solve for x:

1. Set 2cos^2x - 5cosx + 2 = 0: Let's solve this quadratic equation for cosx:

2cos^2x - 5cosx + 2 = 0

This quadratic equation can be factored as:

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

Setting each factor to zero:

2cosx - 1 = 0 or cosx - 2 = 0

2cosx = 1 or cosx = 2

Now, we know that the range of the cosine function is [-1, 1], so there are no real solutions for cosx = 2. However, there is a solution for 2cosx = 1:

2cosx = 1 cosx = 1/2

To find the values of x that satisfy this equation, we can use the inverse cosine function:

x = arccos(1/2)

Using the unit circle or calculator, we find two possible solutions for x:

x = π/3 or x = 5π/3

2. Set √(sinx) = 0: To find the values of x that satisfy this equation, we need to isolate sinx and square both sides (since the square root is involved):

√(sinx) = 0

Square both sides:

sinx = 0

The sine function is zero at multiples of π:

x = 0 or x = π or x = 2π or ...

So the solutions to this part are all the multiples of π.

Now, combining the solutions from both parts, the final set of solutions for the original equation is:

x = π/3, 5π/3, 0, π, 2π, 3π, ...

These are the values of x that satisfy the given equation.