Since [tex] \csc^2{x}=\frac{1}{\sin^2{x}} [/tex] and [tex] \cot^2{x}=\frac{\cos^2{x}}{\sin^2{x}} [/tex], we can rewrite the right side of the equation as
[tex] \frac{1}{\sin^2{x}}-\frac{\cos^2{x}}{\sin^2{x}} =\frac{1-\cos^2{x}}{\sin^2{x}} [/tex]
Using the identity [tex] \sin^2{x}+\cos^2{x}=1 [/tex], we can subtract [tex] \cos^2{x} [/tex] from either side to obtain the identity [tex] \sin^2{x}=1-\cos^2{x} [/tex]
substituting that into our previous expression, the right side of our equation simply becomes
[tex] \frac{\sin^2{x}}{\sin^2{x}}=1 [/tex]
We can now write our whole equation as
[tex] 3\tan^2{x}-2=1 [/tex]
Adding 2 to both sides:
[tex] 3\tan^2{x}=3 [/tex]
dividing both sides by 3:
[tex] \tan^2{x}=1 [/tex]
[tex] \tan{x}=\pm1 [/tex]
When 0 ≤ x ≤ π, tan x can only be equal to 1 when sin x = cos x, which happens at x = π/4, and it can only be equal to -1 when -sin x = cos x, which happens at x = 3π/4
