MHT CET · Maths · Trigonometric Ratios & Identities
The general solution of \(\tan 3 x=1\) is
- A \(x=n \pi, n \in Z\)
- B \(x=n\left(\frac{\pi}{3}\right)+\frac{\pi}{12}, n \in Z\)
- C \(x=n \pi+\frac{\pi}{4}, n \in Z\)
- D \(x=n \pi \pm \frac{\pi}{4}, n \in Z\)
Answer & Solution
Correct Answer
(B) \(x=n\left(\frac{\pi}{3}\right)+\frac{\pi}{12}, n \in Z\)
Step-by-step Solution
Detailed explanation
We know that \(\tan \theta=\tan \alpha\) implies
\(\theta=n \pi+\alpha\), where \(n \in Z\)
Given \(\tan 3 x=1\)
\(\therefore \tan 3 x=\tan \frac{\pi}{4} \Rightarrow 3 x=n \pi+\frac{\pi}{4}\)
\(\therefore \quad \mathrm{x}=\frac{\mathrm{n} \pi}{3}+\frac{\pi}{12}, \quad \mathrm{n} \in Z\)
\(\theta=n \pi+\alpha\), where \(n \in Z\)
Given \(\tan 3 x=1\)
\(\therefore \tan 3 x=\tan \frac{\pi}{4} \Rightarrow 3 x=n \pi+\frac{\pi}{4}\)
\(\therefore \quad \mathrm{x}=\frac{\mathrm{n} \pi}{3}+\frac{\pi}{12}, \quad \mathrm{n} \in Z\)
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