Solving Quadratic and Trigonometric Equations

[abc]

2 questions ...

1- solve :

5 > x^2 >= -9
: x belongs to R
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2- solve :

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

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thanx
regards
abc

 

[marlon]

cos(x+30) = sin(2x) = cos (90-2x)

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

Then solve to x and you are done...
marlon

 

[marlon]

x^2 >= -9 for a real number : this is not possible. You only have 5>x^2>=0. The solution : x < sqrt(5 ) and x>-sqrt(5)

 

[Muzza]

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

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

 

[marlon]

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

 

[HallsofIvy]

marlon: you originally said "x^2 >= -9 for a real number : this is not possible." Muzza's point was that it certainly is possible.

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.

 

[Muzza]

Thank you HallsofIvy ;)

 

[marlon]

marlon: you originally said "x^2 >= -9 for a real number : this is not possible." Muzza's point was that it certainly is possible.

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