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Math Help - Find all solutions in the interval..........

  1. #1
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    Find all solutions in the interval..........

    Find all solutions in the interval -180o ≤ x ≤ 180o

    to the equation cosx + (sin^2)x = 5/4
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  2. #2
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    Quote Originally Posted by sweetG View Post
    Find all solutions in the interval -180o ≤ x ≤ 180o

    to the equation cosx + (sin^2)x = 5/4
    Hello,

    The writing \sin^2 x means (\sin(x))^2

    Remember the identity : \cos^2(x)+\sin^2(x)=1 \implies \sin^2(x)=1-\cos^2(x)

    The equation is now :

    \cos x+1-\cos^2 x=\tfrac 54

    \cos^2 x-\cos x+\tfrac 14=0

    (\cos(x)-\tfrac 12)^2=0

    Solve the final
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  3. #3
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    Does that mean x will equal 60 and - 60 ?

    Is that right and are there any more solutions ?

    I've forgotten how to do these types of questions >_<
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  4. #4
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    Quote Originally Posted by sweetG View Post
    Does that mean x will equal 60 and - 60 ?

    Is that right and are there any more solutions ?

    I've forgotten how to do these types of questions >_<
    Yes, you are correct for the range provided.

    The \cos \theta function has a period of 360^o. If no range was provided, then there would be infinite solution, some of which could be:

    \theta = 300, 420, 660, 780...
    Attached Thumbnails Attached Thumbnails Find all solutions in the interval..........-trig.jpg  
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  5. #5
    Moo
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    Quote Originally Posted by sweetG View Post
    Does that mean x will equal 60 and - 60 ?

    Is that right and are there any more solutions ?

    I've forgotten how to do these types of questions >_<
    That's good

    From the final equation, you have that cos(x)-1/2=0, that is to say cos(x)=1/2

    According to the unit circle there are only 2 values of x that satisfy this, between -180 and 180 (complete turn). It's ok
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  6. #6
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    Thanks Moo and Air
    It makes a lot more sense now
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