The number of values of x∈[0,nπ],n∈I that satisfy log|sinx|(1+cosx)=2 is
0
The equation is meaningful if |sinx|≠0 or1 and 1+cosx≠0
so x≠kπ,k=0,1,…n,and x≠(2k+1)π2,k=0,...n−1.
log|sinx|(1+cosx)=2
⇔1+cosx=|sinx|2=sin2x=1−cos2x⇒(1+cosx)(cosx)=0⇔cosx=0 or cosx=−1⇒cosx=0⇒x=(2k+1)π2.
⇒cosx=−1⇒x=kπ.
So there is no x which satisfy the given equation.