First slide

Trigonometric equations

Question

$T h e v a l u e s o f ‘ b ’ s u c h t h a t t h e e q u a t i o n \frac{b \cos x}{2 \cos 2 x - 1} = \frac{b + \sin x}{(\cos^{2} x - 3 \sin^{2} x) \tan x} p o s s e s s s o l u t i o n s, b e l o n g t o t h e s e t$

Difficult

A

$(- \infty, \frac{1}{2})$
B

$(- \infty, - \frac{1}{2})$
C

$(- \infty, \infty)$
D

$(- \infty, \frac{1}{2}) \cup (1, \infty)$

Solution

$L e t u s f i n d d o m a i n o f g i v e n e q u a t i o n$

$A l s o 2 \cos 2 x - 1 = 2 (\cos^{2} x - \sin^{2} x) - (\cos^{2} x + \sin^{2} x) = \cos^{2} x - 3 \sin^{2} x$
$N o w g i v e n e q u a t i o n r e d u c e s t o b \sin x = b + \sin x$
$\Rightarrow \sin x = \frac{b}{b - 1} \sin c e - 1 \leq \sin x \leq 1,$
$\Rightarrow - 1 \leq \frac{b}{b - 1} \leq 1 \Rightarrow b < \frac{1}{2} o r b > 1 \Rightarrow b \in (- \infty, \frac{1}{2}) \cup (1, \infty)$

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