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Trigonometric Equation

  • Thread starter Icelove
  • Start date
  • #1
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Homework Statement


cos(x)+2tg(x) = 7/(4*cosx(x)) Solve for x on the real numbers set.


Homework Equations


tg(x) = sin(x)/cos(x) ,
cos^2(x) = 1-2*sin^2(x) ,
The determinant is D = b^2 - 4ac ,
Also cos(x) ,tg(x), sin(x) | -1 < x < 1


The Attempt at a Solution


For final arrangement I got is:
-8sin^2(x) + 8sin(x) - 3 = 0

But the determinant is 8^2 - 4*(-8)*(-3) which is 64 - 96 thus the equation doesn't have a real solution.

I checked on WolframAlpha and it has some weird solutions and I was also told that it does have a real solution.
 

Answers and Replies

  • #2
ehild
Homework Helper
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1,802
1.

The Attempt at a Solution


For final arrangement I got is:
-8sin^2(x) + 8sin(x) - 3 = 0



The coefficient 8 in front of sin^2(x) is wrong.

ehild
 
  • #3
16
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Why?
Let's say we rearrange the original equation to: cos^2(x) + 2 sin(x) = 7/4
// We multiplied by cosx(x) //
Then cos^2(x) becomes 1-2sin^2(x)... That multiplied by 4 will be 4-8sin^2(x).
I don't think where I got that wrong.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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Hi Icelove! :smile:

(try using the X2 tag just above the Reply box :wink:)
cos^2(x) = 1-2*sin^2(x)
nooo :redface: … learn your trigonometric identities …

cos2x = 1 - 2sin2x :wink:
 
  • #5
ehild
Homework Helper
15,395
1,802
[tex]\cos{x}\cdot\cos{x}=(\cos{x})^2=1-(\sin{x})^2[/tex]

ehild
 
  • #6
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cos(x)+2tg(x) = 7/(4*cosx(x)) Solve for x on the real numbers set.

By this you the complete equation is

[tex](4 \cdot cos(x)^2)\cdot cos(x) + (4\cdot cos(x)^2)\cdot (2tan(x)) - 7 = 0 [/tex] which implies that

[tex] (4 \cdot cos(x)^2) \cdot (cos(x) + 2 tan(x)) - 7 = 0[/tex]
 
Last edited:
  • #7
16
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ooooooooooooooooohh... Thanks. :D
 

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