Exams › JEE Advanced › Maths

Find the number of solutions of the equation |cot(x)| = cot(x) + 1/sin(x) in the interval x in [0, 2*pi].

1. 0
2. 1
3. 2
4. 3

Correct answer: 1

Solution

When cot(x) >= 0, the equation gives 1/sin(x) = 0, which is impossible. When cot(x) < 0 (i.e., x in (pi/2, pi) union (3*pi/2, 2*pi)), the equation reduces to cos(x) = -1/2. Solutions of cos(x) = -1/2 in [0, 2*pi] are x = 2*pi/3 and x = 4*pi/3. Check: x = 2*pi/3 is in (pi/2, pi) where cot < 0 -> valid. x = 4*pi/3 is in (pi, 3*pi/2) where cot > 0 (not where cot < 0) -> invalid. So only 1 solution: x = 2*pi/3.

Related JEE Advanced Maths questions

• Identify the periodic function among the following:
• Given that α, β, γ, and θ are the smallest positive angles in increasing order for which their sine values equal a positive constant k, what is the result of 4 sin(α/2) + 3 sin(β/2) + 2 sin(γ/2) + sin(θ/2)?
• Consider the function fn(θ) = tan(θ/2)(1 + sec²θ)(1 + sec⁴θ)... (1 + sec²ⁿθ). Which of the following is true?
• Given that (a−b)sin(θ+ϕ) = (a+b)sin(θ−ϕ) and a tan(θ/2) − b tan(ϕ/2) = c, which of the following holds true?
• In a right-angled triangle ABC where ∠B = 90° and the sum of the two legs a and b equals 4, the triangle's area is maximized when ∠C is:
• If the altitudes p₁, p₂, and p₃ of a triangle enclose a circle with a diameter of 4/3 units, what is the minimum possible value of p₁ + p₂ + p₃?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →