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    a) tan(3x+20degrees)=3/2

    b) 2sin^2x+cos^2x=10/9

    I'm unsure on how to work these out. Any help would be appreciated.
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    (Original post by dippers)

    a) tan(3x+20degrees)=3/2

    b) 2sin^2x+cos^2x=10/9

    I'm unsure on how to work these out. Any help would be appreciated.
    For part a), take arctan of both sides, and then solve for x:

    

\displaystyle 3x + 20\degrees = \arctan (3/2)

    For part (b) you need to use the R formula:

    

\displaystyle a \sin \theta + b \cos \theta = R \sin (\theta+\alpha)

    Also, what's the range for the answer meant to be (0 to 180, -180 to 180, 0 to 360...)?
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    (Original post by Qwertish)
    For part a), take arctan of both sides, and then solve for x:

    

\displaystyle 3x + 20\degrees = \arctan (3/2)

    For part (b) you need to use the R formula:

    

\displaystyle a \sin \theta + b \cos \theta = R \sin (\theta+\alpha)

    Also, what's the range for the answer meant to be (0 to 180, -180 to 180, 0 to 360...)?
    Sorry, the range is -90<x<90

    I've figured out how to do the first one I got 12.1 and -25.4
    and i'm unsure on how to do the second one I know i input 1-sin^2(theta) for cos^2(theta) but then i don't know what to do.
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    (Original post by dippers)
    Sorry, the range is -90<x<90

    I've figured out how to do the first one I got 12.1 and -25.4
    and i'm unsure on how to do the second one I know i input 1-sin^2(theta) for cos^2(theta) but then i don't know what to do.
    Oh, sorry, those are sin^2... It's just this:

    

2 \sin^2(x)+\cos^2(x)=10/9

    

2 \sin^2(x)+(1 - \sin^2(x))=10/9

    

\sin^2(x)+1=10/9

    

\sin^2(x)=1/9

    

\sin(x)=1/3
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    (Original post by Qwertish)
    Oh, sorry, those are sin^2... It's just this:

    

2 \sin^2(x)+\cos^2(x)=10/9

    

2 \sin^2(x)+(1 - \sin^2(x))=10/9

    

\sin^2(x)+1=10/9

    

\sin^2(x)=1/9

    

\sin(x)=1/3

    was just about to type this
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    (Original post by Qwertish)
    Oh, sorry, those are sin^2... It's just this:

    

2 \sin^2(x)+\cos^2(x)=10/9

    

2 \sin^2(x)+(1 - \sin^2(x))=10/9

    

\sin^2(x)+1=10/9

    

\sin^2(x)=1/9

    

\sin(x)=1/3
    Thank you so much! It seems simple when I look at it now! And did I get the first one right or were there more solutions?
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    (Original post by dippers)
    Thank you so much! It seems simple when I look at it now! And did I get the first one right or were there more solutions?
    I think there should be 3 solutions (for tan, there is a solution every 180 degrees, and your adjusted range will span 540 degrees).

    I'll check it now (if TenOfThem doesn't beat me to it )


    EDIT:
    Yep, there should be three. One negative and two positive.
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    (Original post by Qwertish)
    (if TenOfThem doesn't beat me to it )
    The rugby is slowing me down
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    (Original post by Qwertish)
    I think there should be 3 solutions (for tan, there is a solution every 180 degrees, and your adjusted range will span 540 degrees).

    I'll check it now (if TenOfThem doesn't beat me to it )


    EDIT:
    Yep, there should be three. One negative and two positive.
    Oh yeah I know now, thanks guys!
 
 
 
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