How do I solve $sin 2 x = - \frac{1}{2}$ $sin 2x=-1/2$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do I solve $sin 2 x = - \frac{1}{2}$ $sin 2x=-1/2$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2017-12-02T16:50:40

$x = \frac{7 π}{12}, \frac{11 π}{12}, \frac{19 π}{12}, \frac{31 π}{12}$ $x=(7pi)/12,(11pi)/12,(19pi)/12,(31pi)/12$

Explanation:

$sin 2 x = - \frac{1}{2}$ $sin2x=-1/2$

$since sin 2 x < 0 then x in third/fourth quadrants$ $"since "sin2x<0" then x in third/fourth quadrants"$

$\Rightarrow 2 x = \frac{π}{6} \leftarrow related acute angle$ $rArr2x=pi/6larrcolor(blue)"related acute angle"$

$note \to 0 \leq 2 x \leq 4 π$ $"note "rarr0<=2x<=4pi$

$2 x = \frac{7 π}{6}, \frac{11 π}{6}, \frac{19 π}{6}, \frac{31 π}{6}$ $2x=(7pi)/6,(11pi)/6,(19pi)/6,(31pi)/6$

$\Rightarrow x = \frac{7 π}{12} or \frac{11 π}{12} or \frac{19 π}{12} or \frac{31 π}{12}$ $rArrx=(7pi)/12" or "(11pi)/12" or "(19pi)/12" or "(31pi)/12$

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How do I solve $sin 2 x = - \frac{1}{2}$ $sin 2x=-1/2$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

1 Answer

Explanation:

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How do I solve sin 2x=-1/2sin2x=−12sin 2x=-1/2 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?

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How do I solve $sin 2 x = - \frac{1}{2}$ $sin 2x=-1/2$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?