Find m for which \(\displaystyle{\sin{{\left({x}\right)}}}+{\cos{{\left({2}{x}\right)}}}={m}\) has solutions,

Vanessa Mccarty

Vanessa Mccarty

Answered question

2022-04-13

Find m for which sin(x)+cos(2x)=m has solutions, no derivatives
Given f:[0,2π]R,f(x)=sin(x)+cos(2x) find the values of m for which the equations f(x)=m has solutions. The problem it is easy solvable with derivatives but it takes some time and careful computations to end up with the solution [2,98]

Answer & Explanation

ktetik7aeg

ktetik7aeg

Beginner2022-04-14Added 10 answers

cos2x=cos2xsin2x=12sin2x , so our expression is
2y2+y+1
for y=sinx. Since 1y1, we are requesting potential quadratic values for y in that range. So, merely think about the related parabola, let's say z=2y2+y+1, in the yz-plane.
I am aware that it is upside-down and that the axis of symmetry is at y=1/4.
z=2(14)2+14+1=98
Then test the boundary values y=±1 to find a minimum value z=-2 at y=-1. By continuity (i.e. the intermediate value theorem) the range is [2,98]

Mdladlaqe2t

Mdladlaqe2t

Beginner2022-04-15Added 7 answers

Make use of the double-angle formula cos(2x)=12sin2(x) multiply by -8 and complete the square. We have
16sin2(x)8sin(x)+1=98m
(4sin(x)1)2=98m
The LHS's minimum value is 0 and its maximum value is 98 for m & the LHS has a maximum value of 25 giving an lower bound of -2 for m. So m[2,98]

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