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(a) If ${ }^{k} P_{2}=72$, find the value of $k$.

(b) Solve the equation : $2 \cos ^{2} \theta-5 \cos \theta=3 ; 0^{\circ} \leq \theta \leq 360^{\circ}$
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(a) $^{k} P_{2}=\frac{k !}{(k-2) !}$ $=\frac{k(k-1)(k-2) !}{(k-2) !}=72$ $k(k-1)=72 \Longrightarrow k^{2}-k-72=0$ $k^{2}-9 k+8 k-72=0 \Longrightarrow(k-9)(k+8)=0$ $k=-8$ or $9 \Longrightarrow k=9$ (since k cannot be negative)
$\text { (b) } 2 \cos ^{2} \theta-5 \cos \theta=3$
\begin{aligned} &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 \Longrightarrow 2 \cos \theta=-1 \\ &\cos \theta=-0.5 \Longrightarrow \theta=\cos ^{-1}(-0.5)=120^{\circ}, 240^{\circ} \end{aligned}
$\cos \theta-3=0 \Longrightarrow \cos \theta=3$ (has no solution).
by Diamond (89,036 points)

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