Trig Graph

f = sin (x/2)
g = cos(x)+1

Let y=x/2

f=sin(y)
g = cos(2y)+1 = 2cos^2(y) - 1 + 1 = 2cos^2(y)

So f=g where sin(y) = 2 cos^2(y)

But I get 102 degrees, not 112.

I get the same answer as engineer

f=sin(x/2) g=1+cos(x)
let y=x/2
f=sin(y) g=1+cos(2y)
set f=g
sin(y)=1+cos(2y)
cos(2y)=2cos²(y)-1 double angle formulae
sin(y)=1+2cos²(y)-1=2cos²(y)
Pythagorean thm sin²(y)+cos²(y)=1
sin(y)=2(sin²(y)-1)
2sin²(y)+sin(y)-2=0
this is a quadratic eqn
let z=sin(y)
2z²+z-2=0
z=[-1±√(1^2-4(2)(-2))]/2(2)=[-1±√sqrt(17)]/4
but |z|≤1 so z≠ [-1-sqrt(17)]/4
so z=[-1+sqrt(17)]/4
y=arcsin(z)=51.337º
&
x=2y=102.667º

Rap

Looks to me like
f(x) = sin(2x)
which gives
-110.353447
as the solution.

Oops, right you are on both the formula for f(x) and on the solution. Pi and -Pi also work, as the diagram shows..

Since
sin(2x) = 2sin(x)cos(x)
it leads directly to the next equation that has to be explained.

markr & engineer, another sloping forehead moment

It does end up in a quartic, but one roots drops out rather quickly when sinx=0 (predicted from the graph). The remaining cubic in sinx has one real and two imaginary roots,and the absolute value of the real root is less than 1.

Solving through I get the same answer as markr--110.353 degrees.

Rap

rap, That should be 1 - sin^2y not sin^y -1 (cos^2(y) + sin^2(y) = 1 so cos^2(y) = 1 - sin^2(y)

never mind, you got the same answer. Ok thanks.