Solve the trigonometric equationSolve the equation cos ⁡ x − 2 cos ⁡ 2 x + 3 cos ⁡ 3 x − 4 cos ⁡ 4 x = 1 2 .I tried, put t = cos ⁡ x

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Solve the trigonometric equationSolve the equation  cos  ⁡  x  −  2  cos  ⁡  2  x  +  3  cos  ⁡  3  x  −  4  cos  ⁡  4  x  =            1      2        .I tried, put   t  =  cos  ⁡  x

barene55d · Accepted Answer

With   c  =  cos  ⁡  x we have   cos  ⁡  2  x  =  2      c    2    −  1, and   cos  ⁡  3  x  =  4      c    3    −  3  c, and finally   cos  ⁡  4  x  =  8      c    4    −  8      c    2    +  1. If you take your equation, move the 1/2 to the left side, make the above substitutions, and multiply by   −  2, you&#039;ll get  64      c    4    −  24      c    3    −  56      c    2    +  16  c  +  5  =  0.This factors as   (  2  c  −  1  )  (  32      c    3    +  4      c    2    −  26  c  −  5  )  =  0. The cubic here has no rational roots, and it looks like one would have to resort to the cubic equation (a mess) to find its zeros. Numerically the zeros of the cubic are about   −  .859  ,  −  .195  ,  +  .929, all can be cosine of an angle. So from the other root   c  =  1      /    2 of the linear factor, you&#039;ll have a total of eight solutions in each interval   [  2  k  π  ,  (  2  k  +  2  )  π  ]. Looks like only the solutions from   cos  ⁡  x  =  1      /    2 will be familiar angles:   π      /    3 and   5  π      /    3 and their translates by   2  π  n

Solve the trigonometric equation Solve the equation cos x -2 cos 2x+3 cos 3x -4 cos 4x = 1/2. I tried, put t=cos x.

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