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    i need help with the question in this link...
    im not sure how to work it out..

    https://imgur.com/a/rIB0A
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    (Original post by Chez 01)
    i need help with the question in this link...
    im not sure how to work it out..

    https://imgur.com/a/rIB0A
    (a) All you need to do is firstly express the equation entirely in terms of \sin \theta, then just use the given substitution x= \sin \theta to obtain what they have.

    (b) (i) That's just a quadratic to solve, I'm sure you can do this.

    (ii) Use the fact that we denoted x = \sin \theta, and along with the solutions to (i) to then solve for \theta in the given domain.
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    (Original post by Chez 01)
    i need help with the question in this link...
    im not sure how to work it out..

    https://imgur.com/a/rIB0A
    3\cos^2 \theta = 3(1-\sin^2 \theta)

    and when you expand that you get

    3 - 3\sin^2 \theta

    You just wrote 3\sin^2\theta in your working.
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    (Original post by RDKGames)
    (a) All you need to do is firstly express the equation entirely in terms of \sin \theta, then just use the given substitution x= \sin \theta to obtain what they have.

    (b) (i) That's just a quadratic to solve, I'm sure you can do this.

    (ii) Use the fact that we denoted x = \sin \theta, and along with the solutions to (i) to then solve for \theta in the given domain.
    thankyou for replying....so i got 90 and 19.5 and 160.5 degrees for the last question...is that right?
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    (Original post by Notnek)
    3\cos^2 \theta = 3(1-\sin^2 \theta)

    and when you expand that you get

    3 - 3\sin^2 \theta

    You just wrote 3\sin^2\theta in your working.
    oh okay...thankyou
 
 
 
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