Please help with equation involving inverse trigonometric functions
Solve for x:
arccos x +2arcsin(√3/2)= π/3
I keep getting 1/2 for my answer but it does not check. Here's my process
arccos x + 2arcsin(√3/2)= π/3
arccos x = π/3 - 2arcsin(√3/2)
x=cos[π/3 - 2arcsin(√3/2)]
Using cosine of a difference identity:
x=cos(π/3)cos2[arcsin(√3/2)] + sin(π/3)sin2[arcsin(√3/2)]
x=(1/2)cos(2π/3) + (√3/2)sin(2π/3)
x=(1/2)(-1/2) + (√3/2)(√3/2)
x= (-1/4) + (3/4)
x=2/4=1/2
Could someone please tell me what I'm doing wrong?
Re: Please help with equation involving inverse trigonometric functions
