Several Mathematics problems?

If anybody would like to tell me the solutions and methods for these math problems I would be very grateful.

1. If sinx= 5/13 , and pi/2 < x < pi , then sin 2x= ???

2. This one might be confusing..------ (10/(x+1)) / (.5 + (3/(x+1)))

3. simplify (x+2) +( (x^2-9x-22)/(x^2-121) )

SOLVE

4. 2cos^2(x)-5cosx-3=0

5. sin2x=cosx

6. simplify--- ((x+y)/x) / (x^-1) + (y^-1)

----- Thanks!

3 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    1. If sin x = 5 / 13 in the second quadrant, then

    (sin x)^2 + (cos x)^2 = 1

    (5 / 13)^2 + (cos x)^2 = 1

    25 / 169 + (cos x)^2 = 1

    (cos x)^2 = 1 - 25 / 169

    (cos x)^2 = 144 / 169

    cos x = +or- sqrt(144 / 169)

    cos x = +or- 12 / 13

    Since pi / 2 < x < pi, then x = -12 / 13.

    If x = -12 / 13 and sin 2x = 2(sin x)(cos x), then

    sin 2x = (5 / 13)(-12 / 13)

    sin 2x = -60 / 169

    Answer: sin 2x = -60 / 169

    2.

    10 / (x + 1)

    --------------------- =

    0.5 + 3 / (x + 1)

    multiply the numerator and denominator by the LCD = x + 1

    10(x + 1) / (x + 1)

    ------------------- ------------------ =

    0.5(x + 1) + 3(x + 1) / (x + 1)

    10

    -------------------- =

    0.5x + 0.5 + 3

    10

    -------------- =

    0.5x + 3.5

    10(10)

    ------------------- =

    10(0.5x + 3.5)

    100

    ---------- =

    5x + 35

    20

    --------

    x + 7

    3. (x + 2) + [ (x + 2)(x - 11) ] / [ (x + 11)(x - 11) ]

    = (x + 2) + (x + 2) / (x + 11)

    = [ (x + 2)( x + 11) + (x + 2) ] / (x + 11)

    = [ x^2 + 13x + 22 + x + 2 ] / (x + 11)

    = [ x^2 + 14x + 24 ] / (x + 11)

    4. 2((cos x)^2) - 5cos x - 3 = 0

    (2cos x + 1)(cos x - 3) = 0

    2cos x + 1 = 0 or cos x - 3 = 0

    cos x = -1 / 2 or cos x = 3 (cos x = 3 has no solution)

    x = 2pi / 3 + 2(pi)k where k is an integer

    or

    x = 4pi / 3 + 2(pi)k where k is an integer

    5. sin 2x = cos x

    2(sin x)(cos x) = cos x

    2(sin x)(cos x) - cos x = 0

    (cos x)(2sin x - 1) = 0

    cos x = 0 or 2sin x - 1 = 0

    cos x = 0 or sin x = 1 / 2

    x = pi / 2 + (pi)k where k is an integer

    or

    x = pi / 6 + 2(pi)k where k is an integer

    or

    x = 5pi / 6 + 2(pi)k where k is an integer

    6. ((x + y) / x) / [ (x^-1) + (y^-1) ] Are you missing some brackets?

    = ((x + y) / x) / [ 1 / x + 1 / y ]

    = ((x + y) / x) / [ (x + y) / (xy) ]

    = ((x + y) / x)( xy / ( x + y))

    = xy / x

    = y

    or ((x + y) / x) / (x^-1) + (y^-1)

    = ((x + y) / x) / ( 1 / x) + (1 / y)

    = ((x + y) / x)( 1 / x) + (1 / y)

    = (x + y) / x^2 + 1 / y

    = [ y(x + y) + (x^2)1 ] / (yx^2)

    = [ x^2 + xy + y^2 ] / (yx^2)

  • Anonymous
    1 decade ago

    1. sin2x = 2sinxcosx

    = 2 x 5/13 x 12/13

    = 120/169

    5. sin2x = cosx

    sin2x - cosx = 0

    2sinxcosx - cosx = 0

    cosx(2sinx - 1) = 0

    cosx = 0 sinx = 1/2

    x = 90,270 x = 30,150

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    4 years ago

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