The maximum value of 1+2sinx+3cos2x is
133
We will first make the expression in terms of sinx only. Then we will try to convert it into perfect square. Once we have it in the form k±a (sinx±b)2, we can easily find its range.
1+2sinx+3cos2x
=1+2sinx+3(1−sin2x)
=1+2sinx+3−3sin2x
=4+2sinx−3sin2x
=4−3(sin2x−23sinx+19−19)
=4−3(sinx−13)2+13
=133−3(sinx−13)2
The expression attains maximum value if (sinx−13)2=0, which is possible if sinx=13.
⇒ Maximum value = 133.