March 30, 2017

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1)Solve sin2x=3cosx for all values of x.

What do I need to do?

2)Solve e(sqrtx)=4

3)Approximate the greatest real root of f(x)=x^3-2x-5 to the nearest tenth.

I just need help not do for me. HELP. Please and Thank you. Explain please no links.

  • Math - ,

    1: sin(2x) = 2sin(x)cos(x), so you know that 2sin(x)cos(x)=3cos(x). So either 2sin(x)=3, or cos(x)=0. You now need to find all the values of x such that one or other of these equations is true.

    2: e^(sqrt(x))=4
    so ln(e^(sqrt(x))=ln(4), but ln(e^y)) = y for any y,
    so sqrt(x)=ln(4),
    so x = (ln(4))^2. Try feeding this value into the original equation, and you should get 4.

    3: Try to find any real root like this:
    If x^3 - 2x - 5 = 0, then x^3 = 2x + 5, so x = (2x+5)^(1/3), i.e. the cube root of (2x+5).
    Try x = 1: feed this into the equation and you'll get (2+5)^(1/3) = 1.913.
    Now feed x=1.913 into the equation and you'll get 2.067.
    Now feed x=2.067 into the equation and you'll get 2.090.
    Now feed x=2.090 into the equation and you'll get 2.094.
    Now feed x=2.094 into the equation and you'll still get 2.094 to 3 places of decimals.
    The series is converging to 2.094, or 2.1 to one place of decimals, so this is one of the roots (though you don't know whether it's the highest one yet).
    If you try it out, you should find that f(2.094)=0. (There are tests to enable you to find out whether the series will converge or not, but it's usually worth a try to see if you get lucky. If you don't get lucky, just try reformulating the equation another way and try that instead.) For x higher than this value, the gradient f'(x) = 3x² - 2 is always positive, so x=2.1 must be the greatest real root.

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