Question

√3sinx+2sin(2x+П/6)=√3sin2x-1 [-3П;-3П/2]

Answer 1

Задание из ЕГЭ, дано с опечаткой. Предлагаю решение уравнения с +1 в конце.

Answer 2

√3*sin x + 2sin(2x+pi/6) = √3*sin 2x - 1

√3*sin x + 2(sin 2x*cos(pi/6) + cos 2x*sin(pi/6)) = √3*sin 2x - 1

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

√3*sin x + √3*sin 2x + cos 2x = √3*sin 2x - 1

Сокращаем слева и справа √3*sin 2x. Преобразуем cos 2x = 1 - 2sin^2 x

√3*sin x + 1 - 2sin^2 x = -1

Переносим всё направо

0 = 2sin^2 x - √3*sin x - 2

Получили квадратное уравнение относительно sin x

D = b^2 - 4ac = 3 - 4*2(-2) = 3 + 16 = 19

sin x1 = (-b - √D)/(2a) = (√3 - √19)/4 ~ -0,65 > -1 - подходит

sin x2 = (-b + √D)/(2a) = (√3 + √19)/4 ~ 1,52 > 1 - не подходит

x1 = arcsin((√3 - √19)/4) + 2pi*n ≈ -0,7165 + 2pi*n - это угол в 4 четверти

x2 = pi - arcsin((√3 - √19)/4) + 2pi*n ≈ 3,858 + 2pi*n - это угол в 3 четверти.

На промежутке (-3pi; -3pi/2) ≈ (-9,42; -4,71) будут корни

x1 = arcsin((√3 - √19)/4) - 2pi ≈ -0.7165 - 2*3,1416 ≈ -7

x2 = pi - arcsin((√3 - √19)/4) - 4pi ≈ 3,858 - 4*3,1416 ≈ -8,708

Answer 3

To solve the equation √3sin(x) + 2sin(2x + π/6) = √3sin(2x) - 1 in the interval [-3π, -3π/2], we need to find the values of x that satisfy the equation in that range.

Let's start by simplifying the equation step by step:

1. Use the trigonometric identity: sin(2x) = 2sin(x)cos(x).
2. Replace sin(2x) in the equation: √3sin(x) + 2sin(2x + π/6) = √3 * sin(x) + 2 * (2sin(x)cos(x) + cos(π/6)) √3sin(x) + 4sin(x)cos(x) + √3/2

Now, our equation becomes: √3sin(x) + 4sin(x)cos(x) + √3/2 = √3sin(2x) - 1

Next, let's bring all the terms to one side of the equation to set it to zero: √3sin(x) + 4sin(x)cos(x) + √3/2 - √3sin(2x) + 1 = 0

Now, we have the equation in the form: f(x) = 0, where: f(x) = √3sin(x) + 4sin(x)cos(x) + √3/2 - √3sin(2x) + 1

To solve this equation, you can use numerical methods or graphing techniques to find the approximate solutions within the given interval [-3π, -3π/2]. Unfortunately, there isn't a straightforward algebraic method to find the exact solutions for this transcendental equation.

For numerical methods, you could use techniques like the Newton-Raphson method or other root-finding algorithms to find the x-values where f(x) is close to zero.

If you have specific values for x within the interval [-3π, -3π/2] and want to check if they satisfy the equation, you can plug them into the equation and verify if the left-hand side equals the right-hand side.