Question

Question #3aca5

Answer 1

`(7pi)/6; (11pi)/6`

Explanation:

1/(sec^2 x) + 3sin (x/2)cos (x/2) = 0
since `1/(sec^2 x) = cos^2 x = 1 - sin^2 x`, and
`3sin (x/2)cos (x/2) = (3/2)sin x`, therefore:
`1 - sin^2 x + (3/2)sin x = 0`
`- 2sin^2 x + 3sin x + 2 = 0`
Solve this quadratic equation for sin x.
`D = d^2 = b^2 - 4ac = 9 + 16 = 25` --> `d = +- 5`
There are 2 real roots:
`sin x = -b/(2a) +- d/(2a) = -3/-4 +- 5/4 = 3/4 +- 5/4`
`sin x = 8/4 = 2` (rejected as > 1), and
`sin x = - 2/4 = - 1/2`
Trig table and unit circle give:
`x = (-pi)/6` or `x = (11pi)/6` (co-terminal), and
`x = pi - (- pi/6) = x + pi/6 = (7pi)/6`
Check by calculator.
`x = (7pi)/6 = 210^@` --> cos x = -0.866 --> `cos^2 x = 0.75` -->
x/2 = 105 --> cos (x/2) = -0.258 --> sin (x/2) = 0.966.
cos^2 x + 3sin (x/2).cos (x/2) = 0.75 + 3(-0.258)(0.966) =
= 0.75 - 0.75 = 0. Correct.