- #1

- 22

- 0

**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

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- Thread starter abc
- Start date

- #1

- 22

- 0

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

- 3,768

- 10

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

Then solve to x and you are done...

marlon

- #3

- 3,768

- 10

- #4

- 695

- 0

x^2 >= -9 for a real number : this is not possible.

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

- #5

- 3,768

- 10

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

- 41,847

- 966

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

- 695

- 0

Thank you HallsofIvy ;)

- #8

- 3,768

- 10

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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