Sin^2x- √3/3*sin2x=cos^2x

Предыдущий
вопрос Следующий
вопрос

To solve this equation, we can start by using the double angle identity for sine:

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

Substitute this into the given equation:

sin^2(x) - √3/3 * 2sin(x)cos(x) = cos^2(x)

Next, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to rewrite the equation:

1 - sin^2(x) - √3/3 * 2sin(x)cos(x) = 1 - cos^2(x)

Rearranging terms:

1 - 1 - sin^2(x) + sin^2(x) - √3/3 * 2sin(x)cos(x) = cos^2(x) - cos^2(x)

-√3/3 * 2sin(x)cos(x) = 0

Multiplying through by 3/2:

-√3 * sin(x)cos(x) = 0

Since the product sin(x)cos(x) can only be 0 when either sin(x) or cos(x) is 0, we have two cases:

1) sin(x) = 0 which implies x = n*π (where n is an integer)
2) cos(x) = 0 which implies x = (2n+1)π/2 (where n is an integer)

Thus, the solutions to the given equation are x = n*π and x = (2n+1)π/2 for all integers n.