To find the solutions to this equation, we need to set each factor equal to 0:
1) sin x + 1 = 0sin x = -1x = arcsin(-1) + 2πnx = -π/2 + 2πn, where n is an integer
2) 2 cos x - 1 = 02 cos x = 1cos x = 1/2x = arccos(1/2) + 2πnx = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer
Therefore, the solutions to the equation (sin x + 1)(2 cos x - 1) = 0 are:x = -π/2 + 2πn, π/3 + 2πn, 5π/3 + 2πn where n is an integer.
To find the solutions to this equation, we need to set each factor equal to 0:
1) sin x + 1 = 0
sin x = -1
x = arcsin(-1) + 2πn
x = -π/2 + 2πn, where n is an integer
2) 2 cos x - 1 = 0
2 cos x = 1
cos x = 1/2
x = arccos(1/2) + 2πn
x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer
Therefore, the solutions to the equation (sin x + 1)(2 cos x - 1) = 0 are:
x = -π/2 + 2πn, π/3 + 2πn, 5π/3 + 2πn where n is an integer.