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anonymous

  • 5 years ago

Solve the following Trig equation. -3-3sin(theta)=cos^2(theta)

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  1. anonymous
    • 5 years ago
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    Re-write cos^2 in terms of sin^2 and you have a quadratic!

  2. anonymous
    • 5 years ago
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    so i need to solve the LHS?

  3. anonymous
    • 5 years ago
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    No. Use the fact cos^2x = 1 - sin^2x, and then move it all to one side and solve as a quadratic

  4. anonymous
    • 5 years ago
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    Im not sure how to do that?

  5. anonymous
    • 5 years ago
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    I'm using x as theta. -3-3sinx = cos^2x => -3-3sinx = 1-sin^2x =>sin^2x - 3sin - 4 = 0

  6. anonymous
    • 5 years ago
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    Ahh i see how you got it now, now how do i set it up to find on all the solutions from 0<x<2pi?

  7. anonymous
    • 5 years ago
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    Just solve it as you would the quadratic y^2 - 3y -4, for sin x. Note you need solutions -1 <= sin(x) <= 1 , you can ignore the others. arcsin this gives you the first solution Generally, if you find one solutions 0<x<pi you can find the other as pi-x , and if the solution is -pi<x<0 , you have to add 2pi first before you can find them. However, in the case of this one (you'll see what I mean) it only gives one solution.

  8. anonymous
    • 5 years ago
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    i have to show my work in radians, how do i get the value to use for 2Kpi

  9. anonymous
    • 5 years ago
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    wut? sin^2x -3sinx - 4 = (sin(x) - 4)(sin(x) + 1) So sin x = ... or ... However -1 =< sin x =< 1 , so we can ignore one of these. And then use sin^-1 to work out the values of x.

  10. anonymous
    • 5 years ago
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    My questtion says solve each of the following trignometric questions on interval 0<x<2pi, express answers for angles as exact values in radians, I've never done that type of problem like you showed me.

  11. anonymous
    • 5 years ago
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    Is the problem that you normally work in degrees, or do you just not know how to find the angle?

  12. anonymous
    • 5 years ago
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    i just dont know how to find the angle on this type of problem

  13. anonymous
    • 5 years ago
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    Oh, OK. Well can you see what the value of sin(x) is from above?

  14. anonymous
    • 5 years ago
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    ... I've got to go, so I'll rush this, sorry. From the factorisation, sin(x) = -1 (we can ignore sin(x) = 4 because it's too big) We use the inverse sin / arcsin / sin^-1 function => arcsin(-1) = -pi/2 But we need a value between 0 and 2pi 20 we can add 2pi => 3/2 pi this is the only value in the inteval.

  15. anonymous
    • 5 years ago
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    Okay thank you for your time!

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