Number of values of x lying in the interval [0, 4π] and satisfying the equation  tan⁡5x+cot⁡3x=0 is

tan⁡5x=−cot⁡3x=tan⁡(π/2+3x)⇒5x=nπ+π2+3x,n∈I⇒2x=(2n+1)π2,n∈I⇒x=(2n+1)π4,n∈I Now, x∈[0,4π]⇒ 0≤(2n+1)π4≤4π⇒ −12≤n≤152⇒0≤n≤7⇒ There are 8 values of x.