The number of solutions of the equation l o g 2 sin ⁡ x ( 1 + cos ⁡ x ) = 2 in the interval 0 , 5 π is

Step-by-step solution

l o g 2 sin ⁡ x 1 + cos ⁡ x = 2 1 + cos ⁡ x = 2 sin ⁡ x 2 = 2 sin 2 ⁡ x 1 + cos ⁡ x = 2 - 2 cos 2 ⁡ x 2 cos 2 ⁡ x + cos ⁡ x - 1 = 0 ⇒ cos ⁡ x + 1 2 cos ⁡ x - 1 = 0 ⇒ cos ⁡ x = 1 2 or cos x = - 1 (rejected) ⇒ x = π 3 , 5 π 3 in 0,2 π But at x = 5 π 3 ,   2 sin ⁡ x is negative ⇒ x = π 3 in 0,2 π ⇒ x = π 3 ,   π 3 + 2 π ,   π 3 + 4 π ⇒ 3