How do you solve $\sqrt{2} sin 2 x = 1$ $sqrt2sin2x=1$ between the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve $\sqrt{2} sin 2 x = 1$ $sqrt2sin2x=1$ between the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-07-19T02:39:45

We have that

$\sqrt{2} sin 2 x = 1 \Rightarrow sin 2 x = \frac{1}{\sqrt{2}} \Rightarrow sin 2 x = \frac{\sqrt{2}}{2} \Rightarrow sin 2 x = sin (\frac{π}{4})$ $sqrt2 sin2x=1=>sin2x=1/sqrt2=>sin2x=sqrt2/2=> sin2x=sin(pi/4)$

Hence from the last relation we get

$2 x = 2 k π + \frac{π}{4} \Rightarrow x = k π + \frac{π}{8}$ $2x=2kpi+pi/4=>x=kpi+pi/8$

or

$2 x = 2 k π + π - \frac{π}{4} \Rightarrow 2 x = 2 k π + 3 \frac{π}{4} \Rightarrow x = k π + 3 \frac{π}{8}$ $2x=2kpi+pi-pi/4=>2x=2kpi+3pi/4=>x=kpi+3pi/8$

Because $0 \leq x \leq 2 π$ $0<=x<=2pi$ acceptable solutions are

$x_{1} = \frac{π}{8}, x_{2} = 9 \frac{π}{8}, x_{3} = 3 \frac{π}{8}, x_{4} = 11 \frac{π}{8}$ $x_1=pi/8,x_2=9pi/8,x_3=3pi/8,x_4=11pi/8$

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How do you solve $\sqrt{2} sin 2 x = 1$ $sqrt2sin2x=1$ between the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve sqrt2sin2x=1√2sin2x=1sqrt2sin2x=1 between the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?

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How do you solve $\sqrt{2} sin 2 x = 1$ $sqrt2sin2x=1$ between the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?