Question

2 sin(x+pi/3)-sqr3cos2x=sin x + sqr3 [-2pi;-pi/2]

Answer 1

Решение дано на фото.

Answer 2

To solve the trigonometric equation 2sin(x + π/3) - √3cos(2x) = sin(x) + √3 over the interval [-2π, -π/2], we need to find the values of x that satisfy the equation within that interval.

Let's begin by simplifying the equation:

2sin(x + π/3) - √3cos(2x) = sin(x) + √3

Using trigonometric identities: sin(x + π/3) = sin(x)cos(π/3) + cos(x)sin(π/3) = (1/2)sin(x) + (√3/2)cos(x) cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x)

Now, substituting these identities into the original equation:

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

Now, let's solve for sin(x):

sin(x) + √3 - √3 + 2√3sin^2(x) = sin(x) + √3

2√3sin^2(x) = 0

sin^2(x) = 0

Now, solve for sin(x):

sin(x) = 0

To find the values of x that satisfy this equation over the interval [-2π, -π/2], we can list the solutions:

1. x = -π
2. x = -3π/2

Both of these solutions lie within the interval [-2π, -π/2].

Therefore, the solutions to the equation 2sin(x + π/3) - √3cos(2x) = sin(x) + √3 over the interval [-2π, -π/2] are x = -π and x = -3π/2.