If e cos 2 x + cos 4 x + cos 6 x + . . . . ∞ log e 2 satisfies the equation t 2 - 9 t + 8 = 0 , then the value of 2 sin x sin x + 3 cos x , where 0 < x < π 2 , is equal to

Step-by-step solution

e cos 2 x + cos 4 x + … … ∞ ln   2 = 2 cos 2 x + cos 4 x + … … ∞ = = 2 cos 2 x 1 - cos 2 x = 2 cot 2 x t 2 - 9 t + 8 = 0 ⇒ t = 1 , 8 ⇒ 2 cot 2 x = 1 ,   8 ⇒ cot 2 x = 0 , 3 0 < x < π 2 ⇒ cot   x = 3 ⇒ 2   sin   x sin   x + 3   cos   x = 2 1 + 3   cot   x = 2 4 = 1 2