Question

2sin^2x=cos^2х-sinx*cosx

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

Let's start by using trigonometric identities to simplify the right-hand side of the equation:

cos^2x - sinx*cosx = cosx(cosx - sinx)

Next, we can substitute this expression back into the original equation and simplify:

2sin^2x = cosx(cosx - sinx)

2(1 - cos^2x) = cosx(cosx - sinx)

Expanding the right-hand side gives:

2 - 2cos^2x = cos^2x - sinx*cosx

Rearranging terms gives:

3cos^2x - sinx*cosx - 2 = 0

We can factor the left-hand side of the equation:

(3cosx - 2)(cosx + 1) = 0

This gives us two possible solutions:

3cosx - 2 = 0, which gives cosx = 2/3

cosx + 1 = 0, which gives cosx = -1

However, we need to check these solutions to make sure they satisfy the original equation:

For cosx = 2/3:

2sin^2x = (2/3)^2 - sinx*(2/3) 2sin^2x + (2/3)sinx - (4/9) = 0

Using the quadratic formula, we find:

sinx = (-2/3 ± sqrt(28))/4

The solutions for sinx are approximately -0.894 and 0.556.

For cosx = -1:

2sin^2x = 1 - sinx*(-1) 2sin^2x + sinx - 1 = 0

Using the quadratic formula, we find:

sinx = (-1 ± sqrt(9))/4

The solutions for sinx are -1 and 1/2.

So, the solutions to the original equation are:

x ≈ 2.286, 4.716, 3π/2, and 7π/6.