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haleyking345

  • 3 years ago

Solving Trig Equations a.) sec x + tan x=1

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  1. haleyking345
    • 3 years ago
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    I don't know what to do.

  2. SithsAndGiggles
    • 3 years ago
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    Recall\[\sec x=\frac{1}{\cos x}\\\tan x = \frac{\sin x}{\cos x}\] \[\sec x+\tan x=1\\ \frac{1}{\cos x}+\frac{\sin x}{\cos x}=1\\ \frac{1+\sin x}{\cos x}=1\\ 1+\sin x=\cos x\] Use the identity \[\sin^2x+\cos^2x=1\] \[1+\sqrt{1-\cos^2x}=\cos x\\ \sqrt{1-\cos^2x}=\cos x-1\\ 1-\cos^2x=(\cos x-1)^2\\ 1-\cos^2x=\cos^2x-2\cos x+1\] Does that help?

  3. haleyking345
    • 3 years ago
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    Would it be cot=1?

  4. SithsAndGiggles
    • 3 years ago
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    Well, that last equation simplifies to \[2\cos^2x-2\cos x=0\] What can you do with that?

  5. haleyking345
    • 3 years ago
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    Do you add 2 cos x to the other side?

  6. haleyking345
    • 3 years ago
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    Then divide by 2

  7. SithsAndGiggles
    • 3 years ago
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    Sure, you could do that, but then you get \[\cos^2x=\cos x\] Then your first instinct would be to divide both sides by a cosx, which would eliminate a potential solution. Here's a hint: What's a common factor of 2cos²x and 2cosx?

  8. haleyking345
    • 3 years ago
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    2? Or is it cos?

  9. SithsAndGiggles
    • 3 years ago
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    Actually, it's both!

  10. SithsAndGiggles
    • 3 years ago
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    Factor out a 2cosx from both terms, and you get \[2\cos x \;(\cos x - 1)=0\] What's next?

  11. haleyking345
    • 3 years ago
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    Add 1 to the other side?

  12. SithsAndGiggles
    • 3 years ago
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    No, that's not it. Think of it this way: When you have \[a\cdot b = 0, \text{ either $a$ or $b$ (or both) must be zero, right?}\] This means that for 2cosx (cosx - 1) to be equal to 0, either \[2\cos x=0 \text{ or } \cos x-1=0\]

  13. haleyking345
    • 3 years ago
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    Oh so whhen they are in paranthesis, you can set them equal to 0?

  14. SithsAndGiggles
    • 3 years ago
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    Parentheses aren't exactly the qualifying factor here, no. When you have the product of two things equal to 0, such as \[(ax+b)(cx+d)=0, \text{ then you can split up the left side and get}\\ ax+b=0 \text{ AND }cx+d=0.\\ \text{Solving both equations gives you $x=-\frac{b}{a}$ and $x=-\frac{c}{d}$ as solutions.}\]

  15. SithsAndGiggles
    • 3 years ago
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    The parentheses are there to show that the term (cosx - 1) is a factor of 0, as is 2cosx.

  16. SithsAndGiggles
    • 3 years ago
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    There's a big difference between \[2\cos x\;(\cos x-1) \text{ and } 2\cos x \cos x -1\]

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