We have sinx⋅tan4x=cosx
⇒sinxcosx⋅tan4x=1
⇒tanx⋅tan4x=1
⇒tanx⋅tan2(2x)=1
⇒2tan2x1−tan22x=1tanx
⇒2(2tanx1−tan2x)1−(2tanx1−tan2x)2=1tanx
⇒2(2tanx1−tan2x)(1−tan2x)2−4tan2x(1−tan2x)2=1tanx
⇒4tanx(1−tan2x)1−2tan2x+tan4x−4tan2x=1tanx
⇒4tanx−4tan3x1−6tan2x+tan4x=1tanx
⇒4tan2x−4tan4x=1−6tan2x+tan4x
⇒5tan4x−10tan2x+1=0
So, tan2x=10±√100−2010=10±4√510
∴tan2x=5−2√55 or tan2x=5+2√55
∴tanx=±(5−2√55)1/2 or tanx=±(5+2√55)1/2
As x∈(−π,2π), tanx has two positive and two negative values.
Also tanx>0∀x∈(−π,−π2)∪(0,π2)∪(π,3π2)
and tanx<0∀x∈(−π2,0)∪(π2,π)∪(3π2,2π)
So the number of values of x which satisfy the given equation is 12.