Question

2sin^2xcosx+1=2sin^2x+cosx

Answer 1

Ответ:

Ответ смотри во вложении

Answer 2

To solve the equation 2sin^2(x)cos(x) + 1 = 2sin^2(x) + cos(x), you can follow these steps:

1. First, move all the terms to one side of the equation to set it equal to zero: 2sin^2(x)cos(x) + 1 - 2sin^2(x) - cos(x) = 0

2. Combine like terms: 2sin^2(x)cos(x) - 2sin^2(x) - cos(x) + 1 = 0

3. Factor out common terms: sin^2(x)(2cos(x) - 2) - cos(x) + 1 = 0

4. Factor out a 2 from the first term: 2sin^2(x)(cos(x) - 1) - cos(x) + 1 = 0

5. Now, you can see that the expression (cos(x) - 1) can be further simplified: 2sin^2(x)(cos(x) - 1) - (cos(x) - 1) = 0

6. Factor out (cos(x) - 1) as a common factor: (cos(x) - 1)(2sin^2(x) - 1) = 0

7. Set each factor equal to zero and solve for x:

   a) cos(x) - 1 = 0 cos(x) = 1

   b) 2sin^2(x) - 1 = 0 2sin^2(x) = 1 sin^2(x) = 1/2

Now, we have two separate equations:

a) cos(x) = 1 b) sin^2(x) = 1/2

Let's solve them individually:

a) For cos(x) = 1: This equation is satisfied when x = 0 (since cos(0) = 1), and it repeats every 2π, so you can write the general solution as: x = 2πn, where n is an integer.

b) For sin^2(x) = 1/2: Take the square root of both sides: sin(x) = ±√(1/2)

The solutions for sin(x) = √(1/2) are x = π/4 + 2πn and x = 7π/4 + 2πn, where n is an integer.

The solutions for sin(x) = -√(1/2) are x = 3π/4 + 2πn and x = 5π/4 + 2πn, where n is an integer.

So, combining all the solutions:

x = 2πn (from the cos(x) = 1 equation) x = π/4 + 2πn x = 7π/4 + 2πn x = 3π/4 + 2πn x = 5π/4 + 2πn

where n is an integer. These are the solutions to the equation 2sin^2(x)cos(x) + 1 = 2sin^2(x) + cos(x).