To find the solutions for the given equations, we will solve them separately.
Cos(3x) * (sin(x) + 1) = 0 The product of two expressions is zero if and only if one or both of the expressions are zero. Thus, we set each factor to zero and solve for x:
Cos(3x) = 0 3x = π/2 + kπ, where k is an integer x = π/6 + (kπ)/3
sin(x) + 1 = 0 sin(x) = -1 x = (3π)/2 + 2kπ, where k is an integer
Therefore, the solutions for the first equation are x = π/6 + (kπ)/3 and x = (3π)/2 + 2kπ.
(tan(2x) + 1)(sin(3x) - √3/2) = 0 As before, the product of two expressions is zero if and only if one or both of the expressions are zero. Thus, we set each factor to zero and solve for x:
tan(2x) + 1 = 0 tan(2x) = -1 2x = π/4 + kπ, where k is an integer x = π/8 + (kπ)/2
sin(3x) - √3/2 = 0 sin(3x) = √3/2 3x = π/3 + 2kπ or 3x = (2π)/3 + 2kπ x = π/9 + (2kπ)/3 or x = (2π)/9 + (2kπ)/3
Therefore, the solutions for the second equation are x = π/8 + (kπ)/2, x = π/9 + (2kπ)/3, and x = (2π)/9 + (2kπ)/3.
To find the solutions for the given equations, we will solve them separately.
Cos(3x) * (sin(x) + 1) = 0
The product of two expressions is zero if and only if one or both of the expressions are zero. Thus, we set each factor to zero and solve for x:
Cos(3x) = 0
3x = π/2 + kπ, where k is an integer
x = π/6 + (kπ)/3
sin(x) + 1 = 0
sin(x) = -1
x = (3π)/2 + 2kπ, where k is an integer
Therefore, the solutions for the first equation are x = π/6 + (kπ)/3 and x = (3π)/2 + 2kπ.
(tan(2x) + 1)(sin(3x) - √3/2) = 0
As before, the product of two expressions is zero if and only if one or both of the expressions are zero. Thus, we set each factor to zero and solve for x:
tan(2x) + 1 = 0
tan(2x) = -1
2x = π/4 + kπ, where k is an integer
x = π/8 + (kπ)/2
sin(3x) - √3/2 = 0
sin(3x) = √3/2
3x = π/3 + 2kπ or 3x = (2π)/3 + 2kπ
x = π/9 + (2kπ)/3 or x = (2π)/9 + (2kπ)/3
Therefore, the solutions for the second equation are x = π/8 + (kπ)/2, x = π/9 + (2kπ)/3, and x = (2π)/9 + (2kπ)/3.