The number of values of x∈0,4π satisfying 3cosx−sinx≥2is

We have x∈0,4π and 3cosx-sinx≥2But 3cosx−sinx≤3+1=2Thus  we  must  have  3cosx−sinx=2 ⇒32cosx−12sinx=1 ⇒cosx+π6=1 ⇒cosx+π6=1,−1⇒                    x+π6=0,2π,4π,6π.....π,3π,5π,....                   ∴x=11π6,23π6 ,5π6,17π6    ∵x∈0,4π