2*cos(x+pi/3)=5 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
2cos(x+3π)=5- this is the simplest trigonometric equation
Divide both parts of the equation by 2
The equation is transformed to
cos(x+3π)=25As 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.
The graph
pi
x1 = - -- + I*im(acos(5/2)) + re(acos(5/2))
3
x1=−3π+re(acos(25))+iim(acos(25))
5*pi
x2 = ---- - I*im(acos(5/2))
3
x2=35π−iim(acos(25))
x2 = 5*pi/3 - i*im(acos(5/2))
Sum and product of roots
[src]
pi 5*pi
- -- + I*im(acos(5/2)) + re(acos(5/2)) + ---- - I*im(acos(5/2))
3 3
(35π−iim(acos(25)))+(−3π+re(acos(25))+iim(acos(25)))
4*pi
---- + re(acos(5/2))
3
re(acos(25))+34π
/ pi \ /5*pi \
|- -- + I*im(acos(5/2)) + re(acos(5/2))|*|---- - I*im(acos(5/2))|
\ 3 / \ 3 /
(35π−iim(acos(25)))(−3π+re(acos(25))+iim(acos(25)))
(5*pi - 3*I*im(acos(5/2)))*(-pi + 3*re(acos(5/2)) + 3*I*im(acos(5/2)))
----------------------------------------------------------------------
9
9(5π−3iim(acos(25)))(−π+3re(acos(25))+3iim(acos(25)))
(5*pi - 3*i*im(acos(5/2)))*(-pi + 3*re(acos(5/2)) + 3*i*im(acos(5/2)))/9
x1 = -1.0471975511966 + 1.56679923697241*i
x2 = 5.23598775598299 - 1.56679923697241*i
x2 = 5.23598775598299 - 1.56679923697241*i