Solve this equation : x 2 + 2 x ⋅ cos ⁡ b ⋅ cos ⁡ c + cos 2 ⁡ b + cos 2 ⁡ c − 1 = 0Such that a + b + c = π I don't have any idea. I can't try anything.

Question

Solve this equation :      x          2        +  2  x  ⋅  cos  ⁡  b  ⋅  cos  ⁡  c  +      cos    2    ⁡      b    +      cos    2    ⁡      c    −  1  =  0Such that   a  +  b  +  c  =  π I don&#039;t have any idea. I can&#039;t try anything.

Selena Pratt · Accepted Answer

&quot;Completing the square&quot; would be my choice.      x    2    +  (  2  c  o  s  (  b  )  c  o  s  (  c  )  )  x  +  c  o      s    2    (  b  )  c  o      s    2    (  c  )  −  c  o      s    2    (  b  )  c  o      s    2    (  c  )  +  c  o      s    2    (  b  )  +  c  o      s    2    (  c  )  −  1  =  0  (  x  +  c  o  s  (  b  )  c  o  s  (  c  )      )    2    =  1  −  c  o      s    2    (  b  )  −  c  o      s    2    (  c  )  +  c  o      s    2    (  b  )  c  o      s    2    (  c  )You say that &quot;a+b+c=π but there is NO &quot;a&quot; in your equation.

dalo2m1ezyxlo · Accepted Answer

x          2        +  2  x  ⋅  cos  ⁡  b  ⋅  cos  ⁡  c  +      cos    2    ⁡      b    +      cos    2    ⁡      c    −  1  =  0      D    4    =      cos    2    ⁡  b  ⋅      cos    2    ⁡  c  −      cos    2    ⁡  b  −      cos    2    ⁡  c  +  1  =      (    1    −          cos      2        ⁡    b    )        (    1    −          cos      2        ⁡    c    )    =      sin    2    ⁡  b  ⋅      sin    2    ⁡  c      x          1      ,      2        =  −  cos  ⁡  b  ⋅  cos  ⁡  c  ±  sin  ⁡  b  ⋅  sin  ⁡  c  =  −  cos  ⁡  (  b  ∓  c  )      x    1    =  −  cos  ⁡  (  b  +  c  )  =  −  cos  ⁡  (  π  −  a  )  =  cos  ⁡  a      x    2    =  −  cos  ⁡  (  b  −  c  )

Solve this equation : x <mrow class="MJX-TeXAtom-ORD"> 2 </mrow> </m

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