cos2θ−6sin‌θ‌cos‌θ+3sin‌2θ+2 ‌=2sin‌2θ−6sin‌θ‌cos‌θ+3 ‌=3−3sin‌2θ+[1−cos‌2‌θ] ‌=4−[3sin‌2θ+cos‌2‌θ] ‌=−√10≤3sin‌2θ+cos‌2‌θ≤√10 For maximum value of given expression 3sin‌2θ+cos‌2‌θ should be maximum Hence maximum value of 4+√10.