Question #3aca5

1 Answer
Jun 16, 2017

7π6;11π6

Explanation:

1/(sec^2 x) + 3sin (x/2)cos (x/2) = 0
since 1sec2x=cos2x=1sin2x, and
3sin(x2)cos(x2)=(32)sinx, therefore:
1sin2x+(32)sinx=0
2sin2x+3sinx+2=0
Solve this quadratic equation for sin x.
D=d2=b24ac=9+16=25 --> d=±5
There are 2 real roots:
sinx=b2a±d2a=34±54=34±54
sinx=84=2 (rejected as > 1), and
sinx=24=12
Trig table and unit circle give:
x=π6 or x=11π6 (co-terminal), and
x=π(π6)=x+π6=7π6
Check by calculator.
x=7π6=210 --> cos x = -0.866 --> cos2x=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.