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## Answers

on finding roots of this quadratic we get cos theta= (root3+-3root3)/8=+-root3

theta= 5pi/6 or 7pi/6. as costheta cant b greater than 1

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2 Sin² Ф + √3 cos Ф + 1 = 0

2 (1- cos²Ф ) + √3 cos Ф + 1 =0

multiply by -1 and rearrange

2 cos² Ф -√3 cos Ф - 3 = 0

Δ = 3 + 24 = 27

cos Ф = ( √3 +- 3 √3 ) / 4 = √3 or -√3 / 2

cos Ф cannot be √3 as it is > 1.

So cos Ф = - √3/2 => Ф = in second and third quadrants. = 5π/6 or 7π/6

or 150 deg or 210 deg

2 (1- cos²Ф ) + √3 cos Ф + 1 =0

multiply by -1 and rearrange

2 cos² Ф -√3 cos Ф - 3 = 0

Δ = 3 + 24 = 27

cos Ф = ( √3 +- 3 √3 ) / 4 = √3 or -√3 / 2

cos Ф cannot be √3 as it is > 1.

So cos Ф = - √3/2 => Ф = in second and third quadrants. = 5π/6 or 7π/6

or 150 deg or 210 deg

**Ф = 2nπ +- 5π/6 or 2nπ +- 7π/6**