Question

2sin^2(2x-п/6)-sin(2x-п/6)-1=0

Answer 1

Решение задания смотри на фотографии

Answer 2

To solve the equation 2sin^2(2x-π/6) - sin(2x-π/6) - 1 = 0, we can use a substitution to simplify the equation. Let's substitute y = sin(2x-π/6):

2y^2 - y - 1 = 0

Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2, b = -1, and c = -1. Plugging in these values:

y = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2)) = (1 ± √(1 + 8)) / 4 = (1 ± √9) / 4 = (1 ± 3) / 4

We have two solutions:

1. y = (1 + 3) / 4 = 4 / 4 = 1
2. y = (1 - 3) / 4 = -2 / 4 = -1/2

Now let's substitute back y = sin(2x-π/6) and solve for x.

For y = 1:

sin(2x-π/6) = 1

2x-π/6 = π/2 + 2πn (n is an integer)

2x = π/2 + π/6 + 2πn

x = (π/2 + π/6 + 2πn)/2

Simplifying the expression:

x = (3π/6 + π/6 + 2πn)/2 = (4π/6 + 2πn)/2 = 2π/3 + πn

For y = -1/2:

sin(2x-π/6) = -1/2

2x-π/6 = -π/6 + 2πm (m is an integer)

2x = -π/6 + π/6 + 2πm

x = ( -π/6 + π/6 + 2πm)/2

Simplifying the expression:

x = (0 + 2πm)/2 = πm

Therefore, the solutions to the equation 2sin^2(2x-π/6) - sin(2x-π/6) - 1 = 0 are:

x = 2π/3 + πn (where n is an integer) x = πm (where m is an integer)