Considering only the space between 𝑥 = 0 𝑎𝑛𝑑 2𝜋, the equation must only have solutions at 𝑥 = 1 and 𝑥 = 2.

Question

Provide a trigonometric equation. Considering only the space between 𝑥 = 0 𝑎𝑛𝑑 2𝜋, the

equation must only have solutions at 𝑥 = 1 and 𝑥 = 2. Explain your thought process and

the work you did to create the equation. You may round decimal values to 3 places.

Answer 1

Here is one way,,

y=sin(((π/7)(x-1))*sin((π/7)(x-2))

set y=0 for equation requested

will have y=0 at x=1 and x=2

y has a half period of 7 so less chance of coming back to zero before x=2π

note, using product to sum trig,

y=(-1/2)[cos((π/7)(2x-3))-cos(π/7])

setting y=0

solutions

x=2+7k ,,,,k=0,+/-1,...

and x=1+7k ,,,,k=0,+/-1,...

Answer 2

sin(kx + b) = 1/2; kx+ b = π/6 or kx + b = 5π/6; k + b = π/6; 2k + b = 5π/6; k = 2π/3; b = π/6 - 4π/6 = - π/2

So, equation sin(2πx/3 - π/2) = 1/2 has only solutions x = 1 and x = 2 on interval [0, 2π].