SATHEE: Trigonometrical Equations 1 Question 9

Trigonometrical Equations 1 Question 9

9. If $5 (\tan^{2} x - \cos^{2} x) = 2 \cos 2 x + 9$ , then the value of $\cos 4 x$ is

(2017 Main)

(a) $- \frac{3}{5}$

(b) $\frac{1}{3}$

(d) $- \frac{7}{9}$

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Answer:

Correct Answer: 9. (d)

Solution:

Given, $5 (\tan^{2} x - \cos^{2} x) = 2 \cos 2 x + 9$

$\Rightarrow 5 \frac{2 \sin^{2} x}{2 \cos^{2} x} - \cos^{2} x = 2 \cos 2 x + 9$

$\Rightarrow 5 \frac{1 - \cos 2 x}{1 + \cos 2 x} - \frac{1 + \cos 2 x}{2} = 2 \cos 2 x + 9$

Put $\cos 2 x = y$ , we have

$\begin{aligned} 5 \frac{1 - y}{1 + y} - \frac{1 + y}{2} = 2 y + 9 \\ \Rightarrow 5 (2 - 2 y - 1 - y^{2} - 2 y) = 2 (1 + y) (2 y + 9) \\ \Rightarrow 5 (1 - 4 y - y^{2}) = 2 (2 y + 9 + 2 y^{2} + 9 y) \\ \Rightarrow 5 - 20 y - 5 y^{2} = 22 y + 18 + 4 y^{2} \\ \Rightarrow 9 y^{2} + 42 y + 13 = 0 \\ \Rightarrow 9 y^{2} + 3 y + 39 y + 13 = 0 \\ \Rightarrow 3 y (3 y + 1) + 13 (3 y + 1) = 0 \\ \Rightarrow (3 y + 1) (3 y + 13) = 0 \\ \Rightarrow y = - \frac{1}{3}, - \frac{13}{3} \\ ∴ \cos 2 x = - \frac{1}{3}, - \frac{13}{3} \\ ∴ \cos 2 x = - \frac{1}{3} ∵ \cos 2 x \neq - \frac{13}{3} \end{aligned}$

$\Rightarrow$	$2 \cos 2 x \cos x + 2 \cos 3 x \cos x = 0$
$\Rightarrow$	$2 \cos x (\cos 2 x + \cos 3 x) = 0$
$\Rightarrow$	$2 \cos x 2 \cos \frac{5 x}{2} \cos \frac{x}{2} = 0$
	$\cos x \cdot \cos \frac{5 x}{2} \cdot \cos \frac{x}{2} = 0$
c	$\begin{aligned} (\cos x & = 0 \cos \frac{5 x}{2} & = 0 \end{aligned}$ )
	$\cos \frac{x}{2} = 0$
ow,	$\cos x = 0$
	$\begin{aligned} (x & = \frac{π}{2}, \frac{3 π}{2} [∵ 0 \leq x < 2 π] \cos \frac{5 x}{2} & = 0 \end{aligned}$ )
$\to$	$\frac{5 x}{2} = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2} \frac{9 π}{2}, \frac{11 π}{2} \dots$
	$x = \frac{π}{5}, \frac{3 π}{5}, π, \frac{7 π}{5}, \frac{9 π}{5} [∵ 0 \leq x < 2 π]$
and	$\cos \frac{x}{2} = 0$
$\Rightarrow$	$\frac{x}{2} = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \dots$
$\Rightarrow$	$[∵ 0 \leq x < 2 π]$
ence,	$x = \frac{π}{2}, \frac{3 π}{2}, π, \frac{π}{5}, \frac{3 π}{5}, \frac{7 π}{5}, \frac{9 π}{5}$