cos(x)=5 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\cos{\left(x \right)} = 5$$
- this is the simplest trigonometric equation
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]
0 + 2*pi - I*im(acos(5)) + I*im(acos(5)) + re(acos(5))
$$\left(0 + \left(2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}\right)\right) + \left(\operatorname{re}{\left(\operatorname{acos}{\left(5 \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}\right)$$
$$\operatorname{re}{\left(\operatorname{acos}{\left(5 \right)}\right)} + 2 \pi$$
1*(2*pi - I*im(acos(5)))*(I*im(acos(5)) + re(acos(5)))
$$1 \cdot \left(2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}\right) \left(\operatorname{re}{\left(\operatorname{acos}{\left(5 \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}\right)$$
(2*pi - I*im(acos(5)))*(I*im(acos(5)) + re(acos(5)))
$$\left(2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}\right) \left(\operatorname{re}{\left(\operatorname{acos}{\left(5 \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}\right)$$
(2*pi - i*im(acos(5)))*(i*im(acos(5)) + re(acos(5)))
x1 = 2*pi - I*im(acos(5))
$$x_{1} = 2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}$$
x2 = I*im(acos(5)) + re(acos(5))
$$x_{2} = \operatorname{re}{\left(\operatorname{acos}{\left(5 \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(5 \right)}\right)}$$
x1 = 6.28318530717959 - 2.29243166956118*i