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