Çözüldü Trigonometric Equation

https://i.ibb.co/BVgzJG9v/equation.png
https://drive.google.com/file/d/11DTYD7voQ9HgZYFJyiNCAnl2WooxmVLn/view
(Page 140, Example 6.8)

Solution - 2:
2·sin(θ) ∓ 3·{ 1 - [ sin(θ) ]^2 }^0,5 = 1
∓3·{ 1 - [ sin(θ) ]^2 }^0,5 = 1 - 2·sin(θ)
9·{ 1 - [ sin(θ) ]^2 } = 1 - 4·sin(θ) + 4·[ sin(θ) ]^2
13·[ sin(θ) ]^2 - 4·sin(θ) - 8 = 0
sin(θ) = (2 / 13)(1 ∓ 3·√3)
n ∈ Z
θ1 = 2π·n + arcsin[ (2 / 13)(1 + 3·√3) ]
θ2 = 2π·n + π - arcsin[ (2 / 13)(1 + 3·√3) ]
θ3 = 2π·n + arcsin[ (2 / 13)(1 - 3·√3) ]
θ4 = 2π·n + π - arcsin[ (2 / 13)(1 - 3·√3) ]

Check by WolframAlpha:
https://www.wolframalpha.com/input?i=13(sin(theta))^2-4sin(theta)-8=0
(Click the "Approximate forms" button)