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anonymous

  • one year ago

Find all solutions to the equation. cos2x + 2 cos x + 1 = 0

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  1. Nnesha
    • one year ago
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    let cosx = x then deal with it as quadratic equation \[\huge\rm x^2 +2x +1 =0\]

  2. Nnesha
    • one year ago
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    can you u factor it ?

  3. anonymous
    • one year ago
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    So it is like (x+1)(x+)1.

  4. anonymous
    • one year ago
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    the 1 should be on the parenthesis*

  5. Nnesha
    • one year ago
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    yes right set it equal to zero

  6. anonymous
    • one year ago
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    It'll be -1.

  7. Nnesha
    • one year ago
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    x = -1 so now we can change x to cos x cos x = -1

  8. anonymous
    • one year ago
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    Is pi the only solution?

  9. Nnesha
    • one year ago
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    yep right or -1 has multiplicity of 2 bec cox = -1 , cosx =- 1 are solution

  10. Nnesha
    • one year ago
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    yes right cos x = -1 = pi

  11. anonymous
    • one year ago
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    Uhm do I have to put pi+2 pi n as the solution or just pi?

  12. Nnesha
    • one year ago
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    pi +2 ?

  13. Nnesha
    • one year ago
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    just pi

  14. Nnesha
    • one year ago
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    \[\pi + 2 \cancel =\pi \]

  15. anonymous
    • one year ago
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    I know that haha, but base on my textbook, "Method for Solving Trigonometric Equations Unless the equation is factorable, use substitution to get just one trigonometric function. Solve the equation for the trigonometric function. Solve for the variable in a general window of +2pin for sine and cosine and +pin for tangent. Find the specific solutions in the given interval by replacing n with integers."

  16. anonymous
    • one year ago
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    So I'm kinda confused.

  17. Mertsj
    • one year ago
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    It said to find all solutions. One solution is pi. Another solution is 3pi. Another solution is 5pi... We write that like this: \[x=\pi+2\pi n\]

  18. anonymous
    • one year ago
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    @Mertsj That's what I'm talking about lol thank you for the help guys!

  19. Mertsj
    • one year ago
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    yw

  20. Nnesha
    • one year ago
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    cool i didn't know this x = pi + 2pi n thingy

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