(a) If (^{k}P_{2} = 72), find the value of k.
(b) Solve the equation : (2cos^{2} theta – 5cos theta = 3; 0° leq theta leq 360°)
Explanation
(a) (^{k}P_{2} = frac{k!}{(k – 2)!})
= (frac{k (k – 1) (k – 2)!}{(k – 2)!} = 72)
(k(k – 1) = 72 implies k^{2} – k – 72 = 0)
(k^{2} – 9k + 8k – 72 = 0 implies (k – 9)(k + 8) = 0)
(k = text{-8 or 9} implies k = 9) (since k cannot be negative).
(b) (2cos^{2} theta – 5cos theta = 3)
(2cos^{2} theta – 5cos theta – 3 = 0)
(2cos^{2} theta – 6cos theta + cos theta – 3 = 0)
(2cos theta (cos theta – 3) + 1(cos theta – 3) = 0)
((2cos theta + 1)(cos theta – 3) = 0)
(2 cos theta + 1 = 0 implies 2cos theta = -1)
(cos theta = -0.5 implies theta = cos^{-1} (-0.5) = 120°, 240°)
(cos theta – 3 = 0 implies cos theta = 3) (has no solution).