(a) If \(^{k}P_{2} = 72\), find the value of k.
(b) Solve the equation : \(2\cos^{2} \theta – 5\cos \theta = 3; 0° \leq \theta \leq 360°\)
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) \(2\cos^{2} \theta - 5\cos \theta = 3\)
\(2\cos^{2} \theta - 5\cos \theta - 3 = 0\)
\(2\cos^{2} \theta - 6\cos \theta + \cos \theta - 3 = 0\)
\(2\cos \theta (\cos \theta - 3) + 1(\cos \theta - 3) = 0\)
\((2\cos \theta + 1)(\cos \theta - 3) = 0\)
\(2 \cos \theta + 1 = 0 \implies 2\cos \theta = -1\)
\(\cos \theta = -0.5 \implies \theta = \cos^{-1} (-0.5) = 120°, 240°\)
\(\cos \theta - 3 = 0 \implies \cos \theta = 3\) (has no solution).