First slide

Trigonometric equations

Question

The values of ‘b’ such that the equation $\frac{b \cos x}{2 \cos 2 x - 1} = \frac{b + \sin x}{(\cos^{2} x - 3 \sin^{2} x) \tan x}$ possess solutions, belong to the set

Moderate

A

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

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

$(- \infty, \infty)$
D

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

Solution

Let us find domain of given equation
$i) 2 \cos 2 x - 1 \neq 0 \Rightarrow x \neq n π \pm \frac{π}{6}$
$i i) \tan x \neq 0 \Rightarrow x \neq n π$
$i i i) \cos^{2} x - 3 \sin^{2} x \neq 0 \Rightarrow x \neq n π \pm \frac{π}{6}$
Also, $2 \cos 2 x - 1 = 2 (\cos^{2} x - \sin^{2} x) - (\cos^{2} x + \sin^{2} x) = \cos^{2} x - 3 \sin^{2} x$
$\frac{b \cos x}{\cos^{2} x - 3 \sin^{2} x} = \frac{b + \sin x}{(\cos^{2} x - 3 \sin^{2} x) \tan x}$
$b \cos x = \frac{b + \sin x}{\frac{\sin x}{\cos x}}$
$bsinx = b + sinx$
$\begin{array}{l} \Rightarrow \sin x = \frac{b}{b - 1} \\ Sin c e - 1 \leq \sin x \leq 1 \end{array}$
$\Rightarrow - 1 \leq \frac{b}{b - 1} \leq 1$
$\Rightarrow \frac{b}{b - 1} + 1 \geq 0 a n d \frac{b}{b - 1} - 1 \leq 0$
$\Rightarrow \frac{2 b - 1}{b - 1} \geq 0 a n d \frac{1}{b - 1} \leq o$
$\Rightarrow b \leq \frac{1}{2} o r b > 1 a n d b < 1$
$\Rightarrow b < \frac{1}{2} o r b > 1$

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