Question

Find all solutions to the equation (sin x)(cos x) = 0 .

a) nπ ∣ n = 0, ± 1, ± 2,
b) π/2 + nπ, nπ/n = 0, ± 1, ± 2,
c) π/2 + 2nπ, nπ/n = 0, ± 1, ± 2,
d) π/2 + nπ, nπ/n = 0, ± 1, ± 2,

Answer 1

The equation (sin x)(cos x) = 0 has solutions when either sin x or cos x is zero. These solutions are nπ, and π/2 + nπ where n is any integer.

Step-by-step explanation:

To find all solutions to the equation (sin x)(cos x) = 0, we need to consider when either sine or cosine is zero, since their product is zero only if at least one of the factors is zero.

• Sin x is zero at nπ, where n is an integer (n = 0, ± 1, ± 2, ...). This corresponds to the horizontal axis crossings on the unit circle.
• Cos x is zero at π/2 + nπ, where n is an integer (n = 0, ± 1, ± 2, ...). This corresponds to the vertical axis crossings on the unit circle.

Therefore, the solutions to the equation are the union of these two sets of solutions:

1. nπ where n = 0, ± 1, ± 2, ...
2. π/2 + nπ where n = 0, ± 1, ± 2, ...

So the correct answer from the provided options would be a combination of (a) and (d), correctly stated as: nπ and π/2 + nπ where n = 0, ± 1, ± 2, ....