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Minchanka [31]
3 years ago
10

PLEASE HELP ME! (80 POINTS UP FOR GRABS!!)

Mathematics
2 answers:
trasher [3.6K]3 years ago
4 0

Answer: A-

Step-by-step explanation:

matrenka [14]3 years ago
3 0

Answer: A

Step-by-step explanation:

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sin(2\theta)=2sin(\theta)cos(\theta)\\\\
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\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0
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sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\
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2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2}
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\measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}

now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).
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