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Marrrta [24]
3 years ago
6

Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Give answers in exact form.

Mathematics
1 answer:
sveticcg [70]3 years ago
7 0

Answer:

(a) x=\frac{5\pi }{3}

(b)  x=\frac{\pi }{8}

Step-by-step explanation:

It is given that 0\leq x\leq 2\pi

(a) We have given equation 2sinx+\sqrt{3}=0

sinx=\frac{-\sqrt{3}}{2}

x=sin^{-1}(-0.866)

x=\frac{-\pi }{3}=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}

(b) tan(2x)=1

2x=tan^{-1}1

2x=\frac{\pi }{4}

x=\frac{\pi }{8}

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Answer:

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Step-by-step explanation:

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6 0
3 years ago
Please help 15 points
strojnjashka [21]

Answer:

2

Step-by-step explanation:

2x(5+4)=(_x5)+(_x4)

lets start with 2x(5+4)

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so we are left with 2x9

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so now we have 18=(_x5)+(_x4)

lets try 2 for the blank spaces

18=(2x5)+(2x4)

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2x4=8

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2x(5+4)=(2x5)+(2x4)

It worked! Hope this helps :)

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