Question 7eb4e

May 19, 2017

$\frac{1}{7} \left(215 - 48 \sqrt{21}\right) \approx - 0.7091$

Explanation:

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

$\therefore 5 {\tan}^{2} x = 7 {\cos}^{2} x + 9 ,$

$\therefore \frac{5 {\sin}^{2} x}{\cos} ^ 2 x = 7 {\cos}^{2} x + 9 ,$

$\therefore 5 {\sin}^{2} x = 5 \left(1 - {\cos}^{2} x\right) = 7 {\cos}^{4} x + 9 {\cos}^{2} x ,$

$\therefore 7 {\cos}^{4} x + 14 {\cos}^{2} x - 5 = 0 ,$

:. cos^2x={-14+-sqrt(14^2-4(7)(-5))}/14, &, because, cos^2x >=0,#

${\cos}^{2} x \ne \frac{- 14 - \sqrt{{14}^{2} - 4 \left(7\right) \left(- 5\right)}}{14} ,$ so that,

${\cos}^{2} x = \frac{- 14 + \sqrt{336}}{14} = \frac{- 14 + 4 \sqrt{21}}{14} ,$ or,

${\cos}^{2} x = - 1 + \frac{2}{7} \sqrt{21.}$

$\therefore 2 {\cos}^{2} x - 1 = 2 \left(- 1 + \frac{2}{7} \sqrt{21}\right) - 1 = - 3 + \frac{4}{7} \sqrt{21.} . . \left(1\right) .$

Now, the Reqd. Value$= \cos 4 x = 2 {\cos}^{2} 2 x - 1 ,$

$= 2 {\left(2 {\cos}^{2} x - 1\right)}^{2} - 1$

$= 2 {\left(- 3 + \frac{4}{7} \sqrt{21}\right)}^{2} - 1 , \ldots \ldots \ldots \ldots \left[\because , \left(1\right)\right]$

$= 2 \left(9 - \frac{24}{7} \sqrt{21} + \frac{48}{7}\right) - 1 ,$

$= \frac{2}{7} \left(111 - 24 \sqrt{21}\right) - 1 ,$

$= \frac{1}{7} \left(215 - 48 \sqrt{21}\right) \approx - 0.7091$

Enjoy Maths.!