If tanx+cotx=2 and x∈[0,100π] then number of values of x =
tanx=1⇒x=nπ+π4=(4n+1)π4,n∈Z
⇒x=π4,5π4,9π4,13π4,..............
Now 0≤(4n+1)π4≤100π ⇒0≤4n+1≤400 ⇒−1≤4n≤399
−14≤n≤99.7