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

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

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Answer 1 · 2016-02-07T00:41:26

We know that

$cos x = 1 - 2 {sin}^{2} (\frac{x}{2})$ $cosx=1-2sin^2(x/2)$

Hence

$sin (\frac{x}{2}) + cos (x) = 0 \Rightarrow sin (\frac{x}{2}) + 1 - 2 {sin}^{2} (\frac{x}{2}) = 0 \Rightarrow 2 {sin}^{2} (\frac{x}{2}) - sin (\frac{x}{2}) - 1 = 0$ $sin(x/2)+cos(x)=0=>sin(x/2)+1-2sin^2(x/2)=0=> 2sin^2(x/2)-sin(x/2)-1=0$

Set $u = sin (\frac{x}{2})$ $u=sin(x/2)$ so we have

$2 u^{2} - u - 1 = 0 \Rightarrow u^{2} - u + u^{2} - 1 = 0 \Rightarrow u (u - 1) + (u - 1) (u + 1) = 0 \Rightarrow (u - 1) (2 u + 1) = 0$ $2u^2-u-1=0=>u^2-u+u^2-1=0=>u(u-1)+(u-1)(u+1)=0=> (u-1)(2u+1)=0$

For $u = 1$ $u=1$ we get $sin (\frac{x}{2}) = 1 \Rightarrow \frac{x}{2} = 2 k π + \frac{π}{2} \Rightarrow x = 4 k π + π$ $sin(x/2)=1=>x/2=2kpi+pi/2=>x=4kpi+pi$

where $k$ $k$ integer.

And for $u = - \frac{1}{2}$ $u=-1/2$ we get $sin (\frac{x}{2}) = - \frac{1}{2} \Rightarrow \frac{x}{2} = 2 k π - \frac{π}{6} \Rightarrow x = 4 k π - \frac{π}{3} or x = 4 k π + 7 \frac{π}{3}$ $sin(x/2)=-1/2=>x/2=2kpi-pi/6=>x=4kpi-pi/3 or x=4kpi+7pi/3$

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

$x = π$ $x=pi$

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

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