We have to solve 5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1
5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1
=> 5(cos x)^2 + 5*( sin x)^2 - 6 tan x*cos x = (cos x)^2 - 1
we know (cos x)^2 + ( sin x)^2 = 1
=> 5 - 6 tan x*cos x = (cos x)^2 - 1
=> 6 - 6 (sin x/ cos x)* cos x = (cos x)^2
=> 6 - 6 sin x = 1 - (sin x)^2
let y = sin x
=> 6 - 6y = 1 - y^2
=> y^2 - 6y+ 5 = 0
=> y^2 - 5y - y + 5 = 0
=> y( y - 5) - 1(y - 5) = 0
=> (y - 1)(y - 5) = 0
y = 1 or 5
As y = sin x, we can ignore x = 5.
So sin x = 1
x = arc sin (1)
x = pi/2 + 2*n*pi
Therefore x = pi/2 + 2*n*pi
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