Start learning →

General solution of cos θ = 0:

Aθ = π/2 + nπ
Bθ = nπ
Cθ = π/4 + nπ
Dθ = 0 + 2nπ
Answer & Solution
Correct answer: A. θ = π/2 + nπ
cos θ = 0 at θ = π/2 and 3π/2 (i.e., 90° and 270°). **General:** θ = π/2 + nπ where n ∈ ℤ. (Every π apart.)
Practise this question with Quill — AP Calculus AB →

Keep practising Unit 3A: Chain Rule and Composite Functions

Browse all Unit 3A: Chain Rule and Composite Functions questions for AP Calculus AB, or explore more Mathematics topics.

Start a Unit 3A: Chain Rule and Composite Functions lesson →
Related questions
In which quadrant is sin θ positive and cos θ negative?If tan θ + cot θ = 2, find sin 2θ:General solution of sin θ = sin α:In triangle with sides 7, 8, 9, find cos of the angle opposite to side 9 (use cosine rule)If sin θ = 12/13 (acute), find tan(θ/2):If sin θ + cos θ = 1, then sin θ × cos θ equals:In triangle ABC with sides a, b, c opposite to angles A, B, C: sine rule states:sin(180° - θ) equals: