Solve the equation in (0, 2pi) using any appropriate method. Answer in radians rounded to four decimal places.
\(\frac{1-sin^2x}{cot^2 x} =\frac{\sqrt{3}}{3}\)
a. 0.8631, 2.2785, 4.0047, 5.4201
b. 0.6888, 2.4528, 3.8304, 5.5944
c. 0.8631, 0.7077, 2.4339, 2.2785
d. 0.6888, 08820, 2.4339, 2.4528
[ 1 - sin^2 (x) ] / cot^2 (x) =
[ cos ^2 (x) ] / [ cos^2 (x) / sin^2(x) ]
sin^2 (x) = sqrt (3) / 3 take both roots
sin (x) = ± sqrt ( 1/ √ 3)
So..... arcsin ( sqrt ( 1/ √ 3) ) ≈ 0.8631 and pi - 0.8631 ≈ 2.2785
And arcsin ( - sqrt ( 1/ √ 3) ) ≈ 5.4201 and pi + 0.8631 ≈ 4.0047
The first answer is correct