Number of values of x lying in the interval [0, 4π] and satisfying the equation tan5x+cot3x=0 is
2
4
6
8
tan5x=−cot3x=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.