MHT CET · Maths · Trigonometric Equations
The general solution of \(\frac{1-\cos 2 x}{1+\cos 2 x}=3\) is
- A \(x=2 n \pi \pm \frac{\pi}{3}, n \in Z\)
- B \(\quad x=n \pi \pm \frac{\pi}{6} \quad, n \in Z\)
- C \(x=2 \mathrm{n} \pi \pm \frac{\pi}{6}, \mathrm{n} \in \mathrm{Z}\)
- D \(x=\mathrm{n} \pi \pm \frac{\pi}{3}, \mathrm{n} \in \mathrm{z}\)
Answer & Solution
Correct Answer
(D) \(x=\mathrm{n} \pi \pm \frac{\pi}{3}, \mathrm{n} \in \mathrm{z}\)
Step-by-step Solution
Detailed explanation
\(\frac{1-\cos 2 x}{1+\cos 2 x}=3 \Rightarrow \frac{2 \sin ^{2} x}{2 \cos ^{2} x}=3\)
\(\therefore \tan ^{2} x=3 \Rightarrow \tan ^{2} x=\tan ^{2} \frac{\pi}{3} \Rightarrow x=n \pi \pm \frac{\pi}{3}\)
\(\therefore \tan ^{2} x=3 \Rightarrow \tan ^{2} x=\tan ^{2} \frac{\pi}{3} \Rightarrow x=n \pi \pm \frac{\pi}{3}\)
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