Posted by Julie on Thursday, April 12, 2012 at 12:35am.
solve for (<_ = less than or equal to / pie = pie sign / pie = negative pie)
3 sin²x = cos²x ; 0 <_ x < 2pie
cos²x  sin²x = sinx ; pie < x <_ pie

Calculus grade 12  Reiny, Thursday, April 12, 2012 at 7:37am
1st one:
3 sin²x = cos²x ; 0 ≤ x < 2π
sin²x /cos²x = 1/3
tan^2 x = 1/3
tanx = ± 1/√3
set your calculator to radians ....
x = .5236
x = π  .5236 = 2.61799
x = π + .5236 = 2.61799
x = 2π .5236 = 5.7596
2nd:
cos^2 x  sin^2 x = sinx
1  sin^2 x  sin^2 x  sinx = 0
2sin^2 x + sinx  1 = 0
(2sinx 1)(sinx + 1) = 0
sinx = 1/2 or sinx = 1
if sinx = 1/2
x = π/6, or x = 5π/6
if sinx = 1
then x = =π/2 or 3π/2 ,but π < x ≤ π
so x = π/2
x = π/2 , π/6 , 5π/5

Calculus grade 12  Julie, Thursday, April 12, 2012 at 11:23am
Thank you Reiny! The first question (1st one) your calculations for the 2nd x is wrong. x = x = π + .5236 = 3.665192654 and not 2.61799

Calculus grade 12  Reiny, Thursday, April 12, 2012 at 1:06pm
yup, clearly a typo, since the first part of it is correct
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