Expert: Sherman D. - 3/24/2008

Question
1) 3 cos x = sec x ; in the interval -pi < x < pi

2) cos(1/2 arcsin (5/13)) find exact value

3) sec x = (-2(sq root of 3)) / 3

Answer
3cos(x) = sec(x)
3cos(x) = 1/cos(x)
3cos(x)^2 = 1
cos(x)^2 = (1/3)
cos(x) = ±sqrt(3)/3
x = about 54.74, 305.26, or about 125.26, 234.74

depending how what you want.

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cos((1/2)sin(5/13)^-1)

lets think of it like this

cos(x/2) = sqrt(((1 + cos(A))/2)

cos((1/2)sin(5/13)^-1) = sqrt(((1 + cos(sin(5/13)^-1)))/2)

sin(5/13)^-1 is the same as saying sin(A) = (5/13)

5^2 + x^2 = 13^2
25 + x^2 = 169
x^2 = 144
x = 12

cos(A) = 12/13

cos((1/2)sin(5/13)^-1) = sqrt(((1 + cos(cos(12/13)^-1))/2)
cos((1/2)sin(5/13)^-1) = sqrt(((1 + (12/13))/2)
cos((1/2)sin(5/13)^-1) = sqrt(((13 + 12)/13)/2)
cos((1/2)sin(5/13)^-1) = sqrt((25/13)/2)
cos((1/2)sin(5/13)^-1 = sqrt(25/26)
cos((1/2)sin(5/13)^-1) = 5sqrt(26)/26

ANS : 5sqrt(26)/26, or .9806

i've already checked the value of 5sqrt(26)/26 and your problem, both match

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3.)
sec(x) = -2sqrt(3)/3
1/cos(x) = -2sqrt(3)/3
cos(x) = 3/(-2sqrt(3))
cos(x) = -3/(sqrt(12))
cos(x) = -3sqrt(12)/12
cos(x) = -6sqrt(3)/12
cos(x) = -sqrt(3)/2
x = 5pi/6, 7pi/6