How do you solve $sin(x/2) + cosx =0$ in the interval [0, 2pi]?

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Answer 1 · 2016-02-07T00:41:26

We know that

$cosx=1-2sin^2(x/2)$

Hence

$sin(x/2)+cos(x)=0=>sin(x/2)+1-2sin^2(x/2)=0=> 2sin^2(x/2)-sin(x/2)-1=0$

Set $u=sin(x/2)$ so we have

$2u^2-u-1=0=>u^2-u+u^2-1=0=>u(u-1)+(u-1)(u+1)=0=> (u-1)(2u+1)=0$

For $u=1$ we get $sin(x/2)=1=>x/2=2kpi+pi/2=>x=4kpi+pi$

where $k$ integer.

And for $u=-1/2$ we get $sin(x/2)=-1/2=>x/2=2kpi-pi/6=>x=4kpi-pi/3 or x=4kpi+7pi/3$

But since our solutions belong to $[0,2pi]$ the acceptable values are

$x=pi$

How do you solve sin(x/2) + cosx =0sin(x/2) + cosx =0 in the interval [0, 2pi]?