The original question was solve for θ in 65 cos ⁡ ( 2 θ ) − 56 sin ⁡ ( 2 θ ) − 55 = 0

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The original question was solve for   θ in   65  cos  ⁡  (  2  θ  )  −  56  sin  ⁡  (  2  θ  )  −  55  =  0

Layla Love · Accepted Answer

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  ≠  0so the substitution is good and doesn&#039;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  =  0that becomes  60      t    2    +  56  t  −  5  =  0and 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  πNote also that your transformation to  cos  ⁡  θ  (  130  cos  ⁡  θ  −  122  sin  ⁡  θ  )  −  10  =  0does not reduce the equation to   130  cos  ⁡  θ  −  122  sin  ⁡  θ  −  10  =  0. From  a  (  b  +  c  )  −  d  =  0you can&#039;t deduce   b  +  c  −  d  =  0

The original question was solve for θ in 65 cos ⁡ (

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