UPSC NDA Math Model Paper 10

Given: 4‌sin‌x+2√3=0 such that x ∈ [0, 2π]
⇒4‌sinx=−2√3
⇒sin‌x=−

√3

2

As we know that, sin x is negative in 3‌rd and 4‌th quadrant.
∵ x ∈ [0, 2π]
⇒ x = 4π/3, 5π/3
Hence, x = 4π/3, 5π/3 are the solution of the given equation.