2cos(x/5)=5 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$2 \cos{\left(\frac{x}{5} \right)} = 5$$
- this is the simplest trigonometric equation
Divide both parts of the equation by 2
The equation is transformed to
$$\cos{\left(\frac{x}{5} \right)} = \frac{5}{2}$$
As right part of the equation
modulo =
True
but cos
can no be more than 1 or less than -1
so the solution of the equation d'not exist.
Sum and product of roots
[src]
10*pi - 5*I*im(acos(5/2)) + 5*I*im(acos(5/2))
$$\left(10 \pi - 5 i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)}\right) + 5 i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)}$$
$$10 \pi$$
(10*pi - 5*I*im(acos(5/2)))*5*I*im(acos(5/2))
$$5 i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)} \left(10 \pi - 5 i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)}\right)$$
25*(2*pi*I + im(acos(5/2)))*im(acos(5/2))
$$25 \left(\operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)} + 2 i \pi\right) \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)}$$
25*(2*pi*i + im(acos(5/2)))*im(acos(5/2))
x1 = 10*pi - 5*I*im(acos(5/2))
$$x_{1} = 10 \pi - 5 i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)}$$
$$x_{2} = 5 i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{5}{2} \right)}\right)}$$
x1 = 31.4159265358979 - 7.83399618486206*i