Solving this equation 10\sin^2\theta-4\sin\theta-5=0 for 0\leq\theta<360^\circ

cooopt392a0i

cooopt392a0i

Answered question

2022-02-27

Solving this equation
10sin2θ4sinθ5=0 for 0θ<360

Answer & Explanation

Hashim Townsend

Hashim Townsend

Beginner2022-02-28Added 5 answers

There is also the purely trigonometric approach:
3cosθ=2sinθ3cosθ+sinθ=2
310cosθ+110sinθ=210
Now setting θ0=arcsin(310) and θ1=arcsin(210), this is equivalent to
sin(θ+θ0)=sinθ1
So the solutions are
θ1θ2+2πZ and πθ1θ2+2πZ
In [0,2π], this leaves
θ1θ0+2π5.72radians and  πθ1θ01.20radians
hence
327,7degrees and  68,7  degrees

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