Question

Cos^2(x-5p/2) -1/2sin2x=0

Answer 1

Вроде так.........................

Answer 2

To solve the trigonometric equation cos^2(x - (5π/2)) - (1/2)sin(2x) = 0, we can use some trigonometric identities and algebraic manipulation.

First, let's simplify the equation step by step:

1. Use the double angle identity for sin(2x):

sin(2x) = 2sin(x)cos(x)

Now, substitute this into the equation:

cos^2(x - (5π/2)) - (1/2)(2sin(x)cos(x)) = 0

2. Simplify further:

cos^2(x - (5π/2)) - sin(x)cos(x) = 0

3. Use the trigonometric identity cos(θ - π/2) = -sin(θ):

(-sin(x))(cos(x)) - sin(x)cos(x) = 0

4. Factor out -sin(x)cos(x):

-sin(x)cos(x)[1 + 1] = 0

5. Combine the terms inside the square brackets:

-sin(x)cos(x)(2) = 0

Now, we have two factors: -sin(x)cos(x) = 0 and 2 = 0.

Let's solve each factor separately:

Factor 1: -sin(x)cos(x) = 0

This factor represents the product of two trigonometric functions. To find its solutions, we can set each factor equal to zero and solve for x separately:

a) -sin(x) = 0 This implies sin(x) = 0.

b) cos(x) = 0

Now, let's find the solutions for each of these equations:

a) sin(x) = 0 x = 0, π, 2π, 3π, ...

b) cos(x) = 0 x = π/2, 3π/2, 5π/2, ...

So, the solutions to the equation -sin(x)cos(x) = 0 are x = 0, π/2, π, 3π/2, 2π, 5π/2, 3π, ...

Factor 2: 2 = 0

This is not a valid equation since 2 is not equal to 0.

Therefore, the solutions to the original equation cos^2(x - (5π/2)) - (1/2)sin(2x) = 0 are x = 0, π/2, π, 3π/2, 2π, 5π/2, 3π, ...