Answer: i) θ = 30°, 60°, 210°, & 240°
ii) θ = 20° & 200°
<u>Step-by-step explanation:</u>
i) sin (2θ) = cos 30°
![\sin(2\theta)=\dfrac{\sqrt3}{2}\\\\.\quad 2\theta=\sin^{-1}\bigg(\dfrac{\sqrt3}{2}\bigg)\\\\.\quad 2\theta=60^o\qquad 2\theta=120^o\\\\.\quad \theta=30^o\qquad \theta=60^o](https://tex.z-dn.net/?f=%5Csin%282%5Ctheta%29%3D%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%5C%5C%5C%5C.%5Cquad%202%5Ctheta%3D%5Csin%5E%7B-1%7D%5Cbigg%28%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%5Cbigg%29%5C%5C%5C%5C.%5Cquad%202%5Ctheta%3D60%5Eo%5Cqquad%202%5Ctheta%3D120%5Eo%5C%5C%5C%5C.%5Cquad%20%5Ctheta%3D30%5Eo%5Cqquad%20%5Ctheta%3D60%5Eo)
To include all of the solutions for one rotation, add 360/2 = 180 to the solutions above. θ = 30°, 60°, 210°, 240°
If you need ALL of the solutions (more than one rotation), add 180n to the solutions. θ = 30° + 180n & 60° + 180n
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ii) cos α = sin (50 + α)
Use the Identity: cos α = sin (90 - α)
Use Transitive Property to get: sin (50° + α) = sin (90° - α)
50° + α = 90° - α
50° + 2α = 90°
2α = 40°
α = 20°
To find all solutions for one rotation, add 360/2 = 180 to the solution above.
α = 20°, 200°
If you need ALL of the solutions (more than one rotation), add 180n to the solution. α = 20° + 180n