If θ ε[-(7 π)/6, (4 π)/3], then the number of solutions of √3 cosec² θ-2(√3-1) cosec θ-4=0, is equal to :

Solution

Maths sir · 3 steps

Treat cosec θ as one unknown, solve the quadratic, then count sine values on the interval, endpoints included.

1. Make it a quadratic. Put x = cosecθ to get √3x² - 2(√3-1)x - 4 = 0. The discriminant is 4(√3-1)² + 16√3 = 16 + 8√3 = 4(√3+1)², a perfect square.

   x = ((√3-1) ± (√3+1))/√3 = 2 or -2/√3

2. Switch to sine. So sin θ = 1/2 or sin θ = -√3/2. Both values lie in [-1, 1], so both give solutions. The interval is [-210°, 240°] in degrees.

   Equation	Solutions in [-210°, 240°]	Count
   sin θ = 1/2	-210° (left endpoint), 30°, 150°	3
   sin θ = -√3/2	-120°, -60°, 240° (right endpoint)	3

3. So. Both endpoints are solutions because the interval is closed, and cosecθ is defined at every one of these angles. There are 3 + 3 = 6 solutions, option (c).

The idea to keep. Turn a quadratic in cosec into values of sin, check each value lies in [−1, 1], then count angles carefully, and always test a closed interval's endpoints.

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