The number of solutions to the equation 2 e sin ⁡ x - 3 - e 2 sin ⁡ x = e sin ⁡ x - e 2 sin ⁡ x - 1 in 0,2 π is

Step-by-step solution

Putting e sin ⁡ x = t , we get, 2 t - 3 - t 2 = t - t 2 - 1 t 2 - 2 t + 3 = t 2 - t + 1 t 2 - 2 t + 3 > 0 , ∀ t ∈ R a n d t 2 - t + 1 > 0 , ∀ t ∈ R a s t h e i r d i s c r i m i n a n t < 0 ⇒ t 2 - 2 t + 3 = t 2 - t + 1 ⇒ t = 2 ⇒ e sin ⁡ x = 2 ⇒ sin ⁡ x = ln ⁡ 2 ∈ 0,1 Hence, from the graph, we get, in 0,2 π there are two solutions