We know that `sin^2(x) + cos^2(x) = 1` . Dividing by `cos^2(x)` gives:

`tan^2x + 1 = sec^2x`

The given equation is : `tan^2(2x) + 5sec^2(2x) = 6 - tan(2x)`` `

`=> tan^2(2x) + 5(1 + tan^2(2x)) = 6 - tan(2x)`

Let `tan(2x) = y`

=> y^2 + 5(1 +...

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We know that `sin^2(x) + cos^2(x) = 1` . Dividing by `cos^2(x)` gives:

`tan^2x + 1 = sec^2x`

The given equation is : `tan^2(2x) + 5sec^2(2x) = 6 - tan(2x)`` `

`=> tan^2(2x) + 5(1 + tan^2(2x)) = 6 - tan(2x)`

Let `tan(2x) = y`

=> y^2 + 5(1 + y^2) = 6 - y

=> y^2 + 5 + 5y^2 = 6 - y

=> 6y^2 + y - 1 = 0

=> 6y^2 + 3y - 2y - 1 = 0

=> 3y(2y + 1) - 1(2y + 1) = 0

=> (3y - 1)(2y + 1) = 0

=> y = 1/3 and y = -1/2

tan 2x = 1/3 and tan 2x = -1/2

2x = arc tan (1/3) and 2x = arc tan(-1/2)

x = 9.21 degrees and x = 166.71 degrees.

**The solution of the equation is x = 9.21 degrees and x = 166.71 degrees.**