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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