Anonymous
Anonymous asked in Social ScienceOther - Social Science · 1 month ago

Solve equation?

1) sin^3(x)+1/8 =0  if (0≤x≤2pi)

2) -4sin^2(x)+4cos(x)+5=0

3) e^cos(x)=1  if x E (-inf, inf)

2 Answers

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  • 1 month ago

    1.

    sin^3(x) = -1/8

    Since x is real, then sin(x) is real, so we have:

    sin(x) = cbrt(-1/8)

    sin(x) = -1/2

    x = (7/6)*pi radians or x = (11/6)*pi radians

    https://www.wolframalpha.com/input/?i=sin%5E3%28x%...

    2.

    -4*sin^2(x) + 4*cos(x) + 5 = 0

    By Pythagoras, sin^2(x) + cos^2(x) = 1, so sin^2(x) = 1 - cos^2(x), so we have:

    -4(1 - cos^2(x)) + 4*cos(x) + 5 = 0

    -4 + 4*cos^2(x) + 4*cos(x) + 5 = 0

    4*cos^2(x) + 4*cos(x) + 1 = 0

    (2*cos(x) + 1)^2 = 0

    2*cos(x) + 1 = 0

    2*cos(x)  = -1

    cos(x) = -1/2

    x = 2*pi*(n +/- (1/3)) radians, for any integer n

    https://www.wolframalpha.com/input/?i=-4*sin%5E2%2...

    3.

    e^cos(x) = 1

    Since x is a real number, then cos(x) is a real number, so we have:

    cos(x) = ln(1)

    cos(x) = 0

    x = pi*(n - 0.5) radians, for any integer n

    https://www.wolframalpha.com/input/?i=e%5Ecos%28x%...

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  • 1 month ago

    1.  Sin^3(x) + 1/8 = 0 --> sin^3(x) = -1/8  take cube root --> sin(x) = -1/2, +1/2

    Note the +1/2 root is degenerate - it can appear twice with a -1/2 root and the the -1/2 root is triple degenerate i.e. (-1/2)^3 = - (1/2)^3 = -1/8

    for the positive root x = pi/4, 3*pi/4

    for the negative root  x = 5*pi/4, 7*pi/4

    2. -4sin^2(x) + 4 cos(x) + 5 = 0  --> use sin^2(x) = 1 - cos^2(x)

    -4 + 4cos^2(x) + 4cos(x) + 5 = 0 -->  4cos^2(x) + 4cos(x) + 1 = 0

    let cos(x) = u --> 4u^2 + 4u +1 = 0 --> u = -2 +/- (1/2)sqrt(16-4) = -2 +/- sqrt(3)

     since u = cos(x)  -1 <= u < = 1 so only +sqrt(3) satisfies this condition

    u = -0.2679 -->  cos(x) = -0.2679  x = 1.842 radians

    3.  e^(cos(x))  = 1

    take ln of both sides  -->  cos(x) = 0 --> x = m*pi/2  where m = all odd negative and positive integers

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