First slide

Trigonometric equations

Question

$If \tan θ = \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}}, where θ \in (0, 2 π), then the number of possible values of θ is$

Moderate

A

1
B

3
C

2
D

5

Solution

$\begin{array}{l} Let \tan θ = x > 0 (\sin c e R H S o f g i v e n e q u a t i o n i s p o s i t i v e) \\ Then x = \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \dots . . \infty}}} \\ \Rightarrow x = \frac{1}{2 + x} \\ \Rightarrow x^{2} + 2 x - 1 = 0 \\ \Rightarrow x = \frac{- 2 \pm \sqrt{8}}{2} = \sqrt{2} - 1 (\sin c e x > 0) \\ ∴ \tan θ = \sqrt{2} - 1 \Rightarrow θ = \frac{π}{8} or \frac{9 π}{8} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App