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2cos2θ−cosθ−1=0 for −180°≤θ≤180° I don't get how it is solved
asked
Apr 6, 2017
Mathematics
middle-school
1
answer
5
votes
131k
views
Simplify. (8a^3 - 3b^2) - (a^3 + b^2)
asked
Feb 3, 2017
Mathematics
high-school
1
answer
5
votes
192k
views
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