- #1

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**2 questions .....**

1- solve :

5 > x^2 >= -9

: x belongs to R

----------------------------------------------

2- solve :

cos (x+30) - sin (2x) = 0

45 >= x >= 0

----------------------------------------------

thanx

regards

abc

- Thread starter abc
- Start date

- #1

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1- solve :

5 > x^2 >= -9

: x belongs to R

----------------------------------------------

2- solve :

cos (x+30) - sin (2x) = 0

45 >= x >= 0

----------------------------------------------

thanx

regards

abc

- #2

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- 9

this means : x+30 = 90-2x and x+30 = -90+2x

Then solve to x and you are done...

marlon

- #3

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- #4

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Huh? It's true for any real number x.x^2 >= -9 for a real number : this is not possible.

- #5

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Muzza said:Huh? It's true for any real number x.

hallo Huh?

I meant that x^2 is always bigger or equal to zero, when x is considered to be a real number. This does not apply for complex numbers though...

regards

marlon

- #6

HallsofIvy

Science Advisor

Homework Helper

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You then noted "I meant that x^2 is always bigger or equal to zero".

Yes: x^2>= 0 which is itself larger than -9. x^2>= -9 for all real numbers x.

If you had said "that's always true" rather than "that's impossible" you would have been right: the solutions to 5>x^2>= -9 is exactly the set of numbers whose square is less than 5.

- #7

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Thank you HallsofIvy ;)

- #8

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HallsofIvy said:

You then noted "I meant that x^2 is always bigger or equal to zero".

Yes: x^2>= 0 which is itself larger than -9. x^2>= -9 for all real numbers x.

If you had said "that's always true" rather than "that's impossible" you would have been right: the solutions to 5>x^2>= -9 is exactly the set of numbers whose square is less than 5.

OK I STAND CORRECTED

regards

marlon

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