If 0 ≤ x ≤ 2 π , then the number of real values of x satisfying the equation 8 1 s i n 2 x + 8 1 c o s 2 x = 30 is

Step-by-step solution

Let 8 1 s i n 2 x = t , then, 8 1 c o s 2 x = 8 1 1 - s i n 2 x = 81 t So, the given equation is t + 81 t = 30 t 2 – 30 t + 81 = 0 ⇒ t = 3 or 27 8 1 s i n 2 x = 3 or 27 ⇒ 3 4 s i n 2 x = 3 1 or 3 3 ⇒ sin 2 ⁡ x = 1 4 or 3 4 ⇒ sin ⁡ x = ± 1 2 , ± 3 2 Hence, there are 8 solutions between 0 and 2 π .