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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