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Posted by on Wednesday, January 12, 2011 at 6:31pm.

how do i solve this equation algebraicly
solve the equation 1+25cos(x)=25sin^2(x)

  • pre-calc - , Wednesday, January 12, 2011 at 7:01pm

    Substitute sin²(x)=1-cos²(x).
    This results in a quadratic equation in cos(x). Solve for y=cos(x) and evaluate x=arccos(y) if -1≤y≤1.

  • pre-calc - , Wednesday, January 12, 2011 at 7:07pm

    1+25cos(x)=25sin^2(x)
    -25 sin^2 x + 25 cos x + 1
    -25 (1 - cos^2 x) + 25 cos x + 1
    -25 + 25 cos^2 x + 25 cos x + 1
    25 cos^2 x + 25 cos x - 24

    substitute u = cos x
    25u^2 + 25u - 24
    complete the square
    u^2 + u - 24/25 = 0
    u^2 + u = 24/25
    u^2 + u + 1/4 = 121/100
    (u + 1/2)^2 = 121/100
    take square root of both sides
    | u + 1/2 | = 11/10
    u + 1/2 = -11/10 and u + 1/2 = 11/10
    u = -8/5 and u = 3/5

    substitute back in u = cos x
    cos x = -8/5 and cos x = 3/5
    x = arccos (-8/5)
    x = arccos (3/5)

  • pre-calc - , Wednesday, January 12, 2011 at 7:15pm

    forgot to put
    I am not a tutor
    I think this is right

  • pre-calc - , Wednesday, January 12, 2011 at 8:22pm

    It's almost perfect.

    In the last step,
    "x = arccos (-8/5)
    x = arccos (3/5) "
    we have to reject the first value of arccos(-8/5) because arccos(-8/5) does not exist.

    For a solution between 0 and 2π, there are two values of x that satisfy arccos(3/5), i.e.
    approx. 53 degrees and 360-53 degrees.
    For the general solution, add 2kπ to the solutions, where k is an integer.

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