Question #73fb6

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Answer 1 · 2017-01-31T03:51:47

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

$\Rightarrow sin (\frac{x}{2}) = 2 {sin}^{2} (\frac{x}{2})$ $=>sin(x/2)=2sin^2(x/2)$

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

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

when $sin (\frac{x}{2}) = 0 = sin 0$ $sin(x/2)=0=sin0$

$=>x/2==npi" where " n in ZZ$

$=>x=2npi$

when $1-2sin(x/2)=0$

$=>sin(x/2)=1/2=sin(pi/6)$

$=>x/2==npi+(-1)^npi/6" where " n in ZZ$

$=>x=2npi+(-1)^npi/3$