Cos3x(sinx+1)=0 (tg2x+1)(sin3x - √3:2)=0

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вопрос

To find the solutions for the given equations, we will solve them separately.

Cos(3x) * (sin(x) + 1) =
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) =
3x = π/2 + kπ, where k is an intege
x = π/6 + (kπ)/3

sin(x) + 1 =
sin(x) = -
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) =
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 =
tan(2x) = -
2x = π/4 + kπ, where k is an intege
x = π/8 + (kπ)/2

sin(3x) - √3/2 =
sin(3x) = √3/
3x = π/3 + 2kπ or 3x = (2π)/3 + 2k
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.