Question #7eb4e

1 Answer
May 19, 2017

Answer:

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

Explanation:

#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.!