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Jaqueline Kirby 2022-06-30 Answered
The original question was solve for θ in 65 cos ( 2 θ ) 56 sin ( 2 θ ) 55 = 0
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Answers (1)

Layla Love
Answered 2022-07-01 Author has 29 answers
Remember the formulas
cos 2 θ = 1 tan 2 θ 1 + tan 2 θ , sin 2 θ = 2 tan θ 1 + tan 2 θ
but first examine the cases θ = π / 2 and θ = π / 2 that would invalidate the substitution.
We have
65 cos π 56 sin π 55 = 120 0 65 cos ( π ) 56 sin ( π ) 55 = 120 0
so the substitution is good and doesn't discard solutions.
Set t = tan θ for simplicity, so you get
65 1 t 2 1 + t 2 56 2 t 1 + t 2 55 = 0
that becomes
60 t 2 + 56 t 5 = 0
and the quadratic has roots
14 + 271 30 14 271 30
so you get
θ = arctan 14 + 271 30 + k π or θ = arctan 14 271 30 + k π
Note also that your transformation to
cos θ ( 130 cos θ 122 sin θ ) 10 = 0
does not reduce the equation to 130 cos θ 122 sin θ 10 = 0. From
a ( b + c ) d = 0
you can't deduce
b + c d = 0
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