Bepaal die algemene oplossing vir elke vergelyking.

       1.1  sin θ = 0,819                         [ A 1.1 ]
       1.2  cos α = 0,337                         [ A 1.2 ]
       1.3  tan θ = 2,605                         [ A 1.3 ]
       1.4  sin α = − 0,468                       [ A 1.4 ]
       1.5  cos Φ = − 0,888                      [ A 1.5 ]
       1.6  tan α = − 0,43                         [ A 1.6 ]
       1.7  cosec α = 2,103                     [ A 1.7 ]
       1.8  sec Φ = 1,257                         [ A 1.8 ]
       1.9  cot θ = − 0,315                        [ A 1.9 ]
       $$ \hspace*{2 mm}\mathrm{1.10\kern3mm\frac{3}{2}\ tan\ α\ =\ \ − 1,077888\kern2mm\ } $$         [ A 1.10 ]
       $$ \hspace*{2 mm}\mathrm{1.11\kern3mm\frac{1}{2}\ cos\ θ\ =\ \ − 0,468\kern2mm\ } $$                 [ A 1.11 ]
       $$ \hspace*{2 mm}\mathrm{1.12\kern3mm\frac{3}{2}\ sin\ θ\ =\ \  0,544\kern2mm\ } $$                     [ A 1.12 ]

       1.13  sin (α + 12,5°)  =  0,819                   [ A 1.13 ]
       1.14  cos (θ − 22,3°)  = − 0,273                   [ A 1.14 ]
       1.15  tan (θ − 32,6°)  =  2,836                    [ A 1.15 ]
       1.16  cot (α + 34,5°) + 1,664 = 0               [ A 1.16 ]
       1.17  2 cosec (θ − 18,3°)  =  4,822             [ A 1.17 ]
       1.18  tan (2θ − 28,4°)  = − 0,430                [ A 1.18 ]

     Los die volgende vergelykings op vir die
     gegewe interval :
       2.1  1,5 tan (θ − 10°)  +  1,077 = 0    en  θ  ∈  [0° ; 180°]
                                                                                [ A 2.1 ]
       2.2  cos 2α  =  sin 12°  en  α  ∈  [0° ; 90°]
                                                                                [ A 2.2 ]
       2.3  4 sin (θ − 15°)  +  2 = 0    en  θ  ∈  [0° ; 270°]
                                                                                [ A 2.3 ]
       2.4  sec (2 Φ + 24,3°)  −  2,5 = 0   en θ  ∈  [− 90° ; 90°]
                                                                                [ A 2.4 ]
       2.5  tan θ − 1  =  − 6°  en  α  ∈  [− 180° ; 180°]
                                                                                [ A 2.5 ]
       2.6  cot (Φ  −  11,8)  =  − 1,664  en  α  ∈  [− 90° ; 270°]
                                                                                [ A 2.6 ]