Question

(8cos^2x-4)*корень квадратный(3*sinx)=0

Answer 1

(8cos^2x-4)*sqrt(3*sinx)=0

Произведение двух множителей равно нулю, если один из множителей равень нулю.

Приравниваем каждый из множителей к нулю

8cos^2x-4=0 (1)

sqrt(3*sinx)=0 (2)

Решаем отдельно, каждое из получившихся равенст

1) 8cos^2x-4=0

Переносим -4 в правую часть уравнения

8cos^2x=4

Делим обе части уравнения на 8 (т.к 8cos^2)

8cos^2x=4|:8

Решаем квадратное уравнение

cos^2x=1/2

cosx1=sqrt2/2

cosx2=-sqrt2/2

x1=+-arccos(sqrt2/2)+2pin, n~Z (значение sqrt2/2 из таблицы = pi/4)

x1=+-pi/4+2pin, n~Z

x2=+-(pi-pi/4)+2pin, n~Z

x2=-+3pi/4+2pin, n~Z

 

2)sqrt(3*sinx)=0

sinx=0

x=pin,n~Z

 

Ответ: x=+-pi/4+2pin, n~Z

x=-+3pi/4+2pin, n~Z

x=pin,n~Z

Answer 2

To solve the equation (8cos^2x - 4) * sqrt(3*sinx) = 0, we can break it down into two separate equations and solve each part individually.

First, let's solve the equation 8cos^2x - 4 = 0:

8cos^2x - 4 = 0 8cos^2x = 4 cos^2x = 4/8 cos^2x = 1/2

Taking the square root of both sides, we have:

cosx = ± sqrt(1/2) cosx = ± sqrt(2)/2

The solutions for cosx are x = π/4 + 2πn, where n is an integer, and x = 7π/4 + 2πn, where n is an integer.

Next, let's solve the equation sqrt(3*sinx) = 0:

sqrt(3*sinx) = 0

To have a square root of zero, the expression inside the square root must be zero. Therefore, sinx = 0.

The solutions for sinx are x = 0 + πn, where n is an integer.

Combining the solutions from both parts, the solutions for the original equation are:

x = π/4 + 2πn, where n is an integer, x = 7π/4 + 2πn, where n is an integer, and x = πn, where n is an integer.