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notka56 [123]
2 years ago
11

Please help me and please

Mathematics
1 answer:
vladimir1956 [14]2 years ago
7 0

Answer:

the answer is 10

Step-by-step explanation:

sorry if im wrong :) :(

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2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3})  } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}

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or~cos\alpha =-\frac{7}{4\sqrt{3} } \\ \alpha =cos^{-1}( \frac{-7}{4\sqrt{3} } )

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3 years ago
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3 years ago
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