MHT CET · Maths · Trigonometric Equations
The solution set of \(8 \cos ^2 \theta+14 \cos \theta+5=0\), in the interval \([0,2 \pi]\), is
- A \(\left\{\frac{\pi}{3}, \frac{2 \pi}{3}\right\}\)
- B \(\left\{\frac{\pi}{3}, \frac{4 \pi}{3}\right\}\)
- C \(\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}\)
- D \(\left\{\frac{2 \pi}{3}, \frac{5 \pi}{3}\right\}\)
Answer & Solution
Correct Answer
(C) \(\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}\)
Step-by-step Solution
Detailed explanation
\(8 \cos ^2 \theta+14 \cos \theta+5=0 \)
\(\therefore 8 \cos ^2 \theta+10 \cos \theta+4 \cos \theta+5=0 \)
\(\therefore 2 \cos \theta(4 \cos \theta+5)+1(4 \cos \theta+5)=0 \)
\(\therefore (2 \cos \theta+1)(4 \cos \theta+5)=0 \)
\(\therefore \cos \theta=\frac{-1}{2} \text { or } \cos \theta=\frac{-5}{4}\)
But \(\cos \theta=\frac{-5}{4}\) is not possible as \(\cos \theta \in[-1,1]\) for all values of \(\theta\).
\(\therefore \cos \theta=\frac{-1}{2} \)
\(\therefore \theta \in\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}\)
\(\therefore 8 \cos ^2 \theta+10 \cos \theta+4 \cos \theta+5=0 \)
\(\therefore 2 \cos \theta(4 \cos \theta+5)+1(4 \cos \theta+5)=0 \)
\(\therefore (2 \cos \theta+1)(4 \cos \theta+5)=0 \)
\(\therefore \cos \theta=\frac{-1}{2} \text { or } \cos \theta=\frac{-5}{4}\)
But \(\cos \theta=\frac{-5}{4}\) is not possible as \(\cos \theta \in[-1,1]\) for all values of \(\theta\).
\(\therefore \cos \theta=\frac{-1}{2} \)
\(\therefore \theta \in\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}\)
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