Question #7eb4e

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Answer 1 · 2017-05-19T13:50:03

$1/7(215-48sqrt21) ~~-0.7091$

$5(tan^2x-cos^2x)=2cos^2x+9,$

$:. 5tan^2x=7cos^2x+9,$

$:. (5sin^2x)/cos^2x=7cos^2x+9,$

$:. 5sin^2x=5(1-cos^2x)=7cos^4x+9cos^2x,$

$:. 7cos^4x+14cos^2x-5=0,$

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

$cos^2x!={-14-sqrt(14^2-4(7)(-5))}/14,$ so that,

$cos^2x=(-14+sqrt336)/14=(-14+4sqrt21)/14,$ or,

$cos^2x=-1+2/7sqrt21.$

$:. 2cos^2x-1=2(-1+2/7sqrt21)-1=-3+4/7sqrt21...(1).$

Now, the Reqd. Value $=cos4x=2cos^2 2x-1,$

$=2(2cos^2x-1)^2-1$

$=2(-3+4/7sqrt21)^2-1,............[because, (1)]$

$=2(9-24/7sqrt21+48/7)-1,$

$=2/7(111-24sqrt21)-1,$

$=1/7(215-48sqrt21) ~~-0.7091$

Enjoy Maths.!