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Dmitry_Shevchenko [17]
2 years ago
12

Solve 2csc^2x-2cscx-1=0 for 0° ≤ x ≤ 360°

Mathematics
1 answer:
umka21 [38]2 years ago
6 0

Answer:

Step-by-step explanation:

2 csc²x-2 csc x-1=0

or

\frac{2}{sin^2x} -\frac{2}{sin x} -1=0\\multiply~by~sin^2x\\2-2sin~x-sin^2x=0\\sin^2x+2sinx-2=0\\sin ~x=\frac{-2 \pm\sqrt{2^2-4*1*(-2)} }{2*1} \\=\frac{-2 \pm\sqrt{4+8} }{2} \\=\frac{-2 \pm\sqrt{4*3} }{2} \\=\frac{-2 \pm2\sqrt{3} }{2} \\=-1\pm\sqrt{3} \\|sin~x| \le ~1\\so~sin~x=-1+\sqrt{3} \\sin~x=\sqrt{3} -1\\x=sin^{-1}(\sqrt{3} -1) \approx 47.06^\circ,180-47.06\\or~x=47.06^\circ,132.94^\circ

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Step-by-step explanation:

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