Solve the equation : \(2 \cos ^{2} \theta-5 \cos \theta=3 ; 0^{\circ} \leq \theta \leq 360^{\circ}\)

Question

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

(b) Solve the equation : \(2 \cos ^{2} \theta-5 \cos \theta=3 ; 0^{\circ} \leq \theta \leq 360^{\circ}\)

Answer 1

(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).