Question

Hi there,
Can someone help me with the question:

"Prove that the value of 7sinx +3cos2x cannot be greater than 5/1/24 for all values of x between 0 degrees and 360 degrees"

Thanks lots. =)

I've already simplified 7sinx + 3cos2x into (2sinx-3)(3sinx+1).

Answer 1

I do not agree with your factoring
7 sin x +3 (1 - 2sin^2x)
7 sin x + 3 - 6 sin^2 x
-[6 sin^2 x - 7sin x -3 )
to find the maximum or minimum take the derivative and set to 0
-[ 12 sin x cos x - 7 cos x ] = 0
sin x = 7/12
then
sin^2 x = 49/144
7 sin x = 49/12
2 sin^2 x = 98/144
so function = 49/12 +3 (1 - 98/144)
= 588/144 +432/144 -98*3/144
= 726/144 = 5.041666...
1/24 = .04166666..
so
5 1/24 is right

Answer 2

Oh, I do agree with the factoring but did not do it.

Answer 3

Here without using calculus
let sin x = z
-y = 6z^2-7z -3
that is a parabola
complete the square to find the vertex
-y/6 = z^2 -7/6 z -1/2
-y/6 +1/2 = z^2 -7/6 z
add (7/12)^2 or 49/144 to both sides
-y/6 +72/144 +49/144 = z^2 -7/6 z + 49/144
-y/6 + 121/144 = (z-7/12)^2
so

vertex is where z = sin x = 7/12

then continue as I did with the calculus way

Answer 4

Thanks for the help =) Clear and detailed explanation.