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Solve sin(θ) = √2/2 for all solutions. Use n as an arbitrary integer.

a) θ = π/4 + nπ
b) θ = 3π/4 + nπ
c) θ = 5π/4 + nπ
d) θ = 7π/4 + nπ

User The Roy
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Final answer:

The equation sin(θ) = √2/2 is solved for all solutions by finding the angles where the sine function equals √2/2 on the unit circle, which are θ = π/4 and θ = 3π/4, and then adding multiples of the period 2π for each solution.

Step-by-step explanation:

To solve the equation sin(θ) = √2/2 for all solutions, we need to consider the unit circle where the sine function has the value of √2/2. This occurs at two principal angles in the unit circle, which are θ = π/4 and θ = 3π/4. Since sine has a period of 2π, the general solutions for θ where sin(θ) equals √2/2 are given by adding integer multiples of 2π to these principal angles.

The correct answers for the equation sin(θ) = √2/2 are therefore:

  • θ = π/4 + 2nπ
  • θ = 3π/4 + 2nπ

Where n is an arbitrary integer. This allows us to include all possible solutions that lie in different cycles of the sine function's periodic behavior.

User Pioz
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