Question

Anybody know the correct answer?

Answer 1

Since $\csc^2{x}=\frac{1}{\sin^2{x}}$ and $\cot^2{x}=\frac{\cos^2{x}}{\sin^2{x}}$, we can rewrite the right side of the equation as

$\frac{1}{\sin^2{x}}-\frac{\cos^2{x}}{\sin^2{x}} =\frac{1-\cos^2{x}}{\sin^2{x}}$

Using the identity $\sin^2{x}+\cos^2{x}=1$, we can subtract $\cos^2{x}$ from either side to obtain the identity $\sin^2{x}=1-\cos^2{x}$

substituting that into our previous expression, the right side of our equation simply becomes

$\frac{\sin^2{x}}{\sin^2{x}}=1$

We can now write our whole equation as

$3\tan^2{x}-2=1$

Adding 2 to both sides:

$3\tan^2{x}=3$

dividing both sides by 3:

$\tan^2{x}=1$

$\tan{x}=\pm1$

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