Q) If satisfies the equation x² - 28x + 27 = 0, find the value of . Given that x lies between 0 and pi/2 .

A) The given expression can be simply rewritten as:

= 3^ (sin^2 x + sin^4 x + sin^6 x ....infinity)

= 3^[ (sin^2 x) / (1 - sin^2x)] (Using concept of sum of infinite GP)

= 3^(tan^2 x)


Now,

x^2 - 28x + 27 = 0

x^2 - 27x - x + 27 = 0

(x-1)(x-27) = 0

roots are 1, 27

Given that 3^(tan^2 x) satisfies the equation so it = 1, 27

3 ^(tan^2 x) = 1, 27

tan^2 x = 0, 3

tan x = 0, +- rt(3)

x = 0, +- pi/3

Since for second part of the question x lies between 0 and pi/2, so we take x = pi/3
So
1/ (cosx + sinx) = 1 / (cos60 + sin60)

= 1/ (1/2 + sqrt(3)/2)

= 2 / ( 1 + sqrt(3) )

________________________________________________________