How do you solve $sin x = - \frac{\sqrt{2}}{2}$ $sin x=-sqrt2/2$ ?

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How do you solve $sin x = - \frac{\sqrt{2}}{2}$ $sin x=-sqrt2/2$ ?

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Answer 1 · 2018-03-06T11:56:30

$see explanation$ $"see explanation"$

Explanation:

$since sin x < 0 then x is in third/fourth quadrant$ $"since "sinx<0" then x is in third/fourth quadrant"$

$\Rightarrow x = {sin}^{- 1} (\frac{\sqrt{2}}{2}) = \frac{π}{4} \leftarrow related acute angle$ $rArrx=sin^-1(sqrt2/2)=pi/4larrcolor(red)"related acute angle"$

$\Rightarrow x = π + \frac{π}{4} = \frac{5 π}{4} \leftarrow in third quadrant$ $rArrx=pi+pi/4=(5pi)/4larrcolor(red)"in third quadrant"$

$\Rightarrow x = 2 π - \frac{π}{4} = \frac{7 π}{4} \leftarrow in fourth quadrant$ $rArrx=2pi-pi/4=(7pi)/4larrcolor(red)"in fourth quadrant"$

$since sin is periodic these solutions will be repeated$ $"since sin is periodic these solutions will be repeated"$
$every 2 π$ $"every"2pi$

$rArrx=(5pi)/4+2kpitok inZZ$

$rArrx=(7pi)/4+2kpitok inZZ$

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How do you solve $sin x = - \frac{\sqrt{2}}{2}$ $sin x=-sqrt2/2$ ?

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