The following equation cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval​ [0,2π​).

Question

The following equation cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval​ [0,2π​).

4sin2x-x^3+1=0

Hi, I understand that I have to set the equation as y=4sin2x-x^3+1 and then entered into the graphing calculator, but I think I've putting it on with the wrong format. I would really appreciate if someone can explain to me the correct way to enter it to y= on the calculator (TI-84 Plus).

Thanks!

Answer 1

is it sin(2x) or (sin(x))^2 ?

I am asking if the 2 is part of the sine argument or is it exponent?

Assuming the former.... y = f(x) = 4 sin(2x) - x^3+1

the 3 approximate solutions are -1.1974, -0.1266, 0.3694

Assuming the latter... y = f(x) = 4 (sin(x))^2 - x^3 + 1

the only approximate solution is 1.702

Per the function plotter at Desmos dot com

Here is a link to the manual for your calculator, written so that it won't get flagged:

https: // www DOT manualslib DOT com / manual / 325940 / Texas-Instruments-Ti-84-Plus DOT html