The number of solutions of the equation
|cot x| = cot x + 1/(sin x) in the interval [ 0, 2π ] is

Solution

Maths sir · 5 steps

Split on the sign of cot x, solve each case, then check every root against the sign you assumed.

1. Domain first. Both cot x and 1/(sin x) need sin x ≠ 0, so x = 0, π, 2π are not allowed.
2. First case. If cot x ≥ 0, then |cot x| = cot x, so the equation becomes 1/(sin x) = 0. That never happens, so this case gives no solution.
3. Second case. If cot x < 0, then |cot x| = -cot x, so -cot x = cot x + 1/(sin x). Multiply through by sin x (which is not zero) to get -2 cos x = 1.

   cos x = -1/2 ⇒ x = 2π/3 or 4π/3

4. Check the sign. At x = 2π/3, cos x < 0 and sin x > 0, so cot x < 0, as this case needs. At x = 4π/3, both are negative, so cot x > 0 and this root is rejected.
5. So. Only x = 2π/3 works. Check: |cot x| = 1/√3 and -1/√3 + 2/√3 = 1/√3. The number of solutions is 1.

The common slip: 2. Counting both roots of cos x = -1/2 gives 2, but at x = 4π/3 we have cot x > 0, which breaks the case cot x < 0.

The idea to keep. With a modulus, every case comes with a condition. After solving inside a case, always test each root against that condition before you count it.

Answer from the NTA final answer key. Solution by PrepWiser.