Question

slove 1+cos^2(x/2)-sinx = 0

Answer 1

\[1+\cos ^{2} (\frac{ x }{ 2 })-\sin x=0 \]
Is this the equation?

Answer 2

@coolme

Answer 3

Ok, then 1+ cos ^2 (x/2)= sin x. But the maximum value of sin x is 1 and cos ^2 (x/2) >=0
so the only possibility is that cos ^2 (x/2)=0, sin x=1 i.e. cos (x/2)=0, sin x=1
i.e. x/2= 2n pi + pi/2 and x= n pi+ (-1)^n (pi/2)
i.e. x=npi +(pi/4) and x= n pi +(-1)^n (pi/2) which is not valid for any value of n, so the given system has no solution

Answer 4

here is a longer way to show this
reccall the half angle property
\[\cos^2(x/2)=\frac{ 1+\cos(x) }{ 2 }\]

\[1 + \frac{ 1+\cos(x) }{ 2 }-\sin(x)=0\]

you can use sin^2 x + cos^2 x = 1, to get
\[\sin(x)=\sqrt{1-\cos^2(x)}\]

\[1+\frac{ 1+\cos(x) }{ 2 }-\sqrt{1-\cos^2(x)}=0\]
\[\large [\frac{ 3+\cos(x) }{ 2 }]^2 = 1-\cos^2(x)\]
\[\frac{ 1 }{ 4 }\cos^2(x)+\frac{ 3 }{ 2 }\cos(x)+\frac{ 9 }{ 4 }=1-\cos^2(x)\]
\[\large \cos^2(x)+10\cos(x)+5=0\]

That has a quadratic form of Ax^2+Bx+C=0, let cos(x)=u

u^2 + 10u + 5=0
no real solutions to that, \[\sqrt{b^2-4*a*c}=\sqrt{6^2-4*5*5=}\sqrt{-64}\]