Solving a sin ⁡ x + b cos ⁡ x = cSuppose a sin ⁡ x + b cos ⁡ x = c . My teacher told me that with changing variable we can solve it by this : ( c + b ) t 2 − 2 a t + ( c − b ) = 0 where t = tan ⁡ x 2 . I thought it is true always but today found a problem . When we define sin ⁡ x and cos ⁡ x with tan ⁡ x 2 , we should consider cos ⁡ x 2 ≠ 0 and x ≠ 2 k π + π but it can be the answer of the main equation (i.e. a sin ⁡ x + b cos ⁡ x = c ) ! So , that formula is incomplete ?

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Solving   a  sin  ⁡  x  +  b  cos  ⁡  x  =  cSuppose   a  sin  ⁡  x  +  b  cos  ⁡  x  =  c . My teacher told me that with changing variable we can solve it by this :   (  c  +  b  )      t    2    −  2  a  t  +  (  c  −  b  )  =  0 where   t  =  tan  ⁡      x    2   . I thought it is true always but today found a problem . When we define   sin  ⁡  x and   cos  ⁡  x with   tan  ⁡      x    2   , we should consider   cos  ⁡      x    2    ≠  0 and   x  ≠  2  k  π  +  π but it can be the answer of the main equation (i.e.   a  sin  ⁡  x  +  b  cos  ⁡  x  =  c ) ! So , that formula is incomplete ?

Maggie Bowman · Accepted Answer

use that            a      2        +          b      2            (          a                        a          2                +                  b          2                      sin    ⁡    (    x    )    +          b                        a          2                +                  b          2                      cos    ⁡    (    x    )    )    =  cand  cos  ⁡  (  ϕ  )  =      a                  a        2            +              b        2              sin  ⁡  (  ϕ  )  =      b                  a        2            +              b        2            therefore we get  sin  ⁡  (  x  +  ϕ  )  =      c                  a        2            +              b        2            or you write  2                      a        tan        ⁡                  (          x                      /                    2          )                            1        +                              (            tan            ⁡                          (              x                              /                            2              )                        )                                2                                +                    b                  (          1          −                                    (              tan              ⁡                              (                x                                  /                                2                )                            )                                      2                                )                            1        +                              (            tan            ⁡                          (              x                              /                            2              )                        )                                2                                =  cwith  tan  ⁡  (  x      /    2  )  =  tas your teacher said

Solving a sin ⁡ x + b cos ⁡ x = c Suppose

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