if sec4x-sec2x=2 find the value of cos^2(x) , 0<x<90?

Answer 1

Sec[2x] = 1/(2Cos²(x) - 1)
Sec[4x] = 1/(2((2Cos²(x) - 1)² - 1)
Let z = Cos²(x)
Then we have:
1/(2(2z - 1)² - 1) - 1/(2z - 1) = 2
which eventually reduces to the quadratic equation:
16z² - 20z + 5 = 0
with solutions z = (1/8)(5 ± √5)