Solution - Equations reducible to quadratic form

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "s4"   was replaced by   "s^4".  1 more similar replacement(s).

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                c*o*s*x+c*o*s^7*x-(c*o*s^4*x)=0 

Step by step solution :

Step  1  :

Step  2  :

Pulling out like terms :

 2.1     Pull out like factors :

   cos⁷x - cos⁴x + cosx  =   cosx • (s⁶ - s³ + 1) 

Trying to factor by splitting the middle term

 2.2     Factoring  s⁶ - s³ + 1 

The first term is,  s⁶  its coefficient is  1 .
The middle term is,  -s³  its coefficient is  -1 .
The last term, "the constant", is  +1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1 

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -1 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  2  :

  cosx • (s6 - s3 + 1)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 3.2     Solve   cosx  = 0

Setting any of the variables to zero solves the equation:

 c  =  0
 o  =  0
 s  =  0
 x  =  0

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 3.3     Solve   s⁶-s³+1 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  r , such that  r = s³  transforms the equation into :
 r²-r+1 = 0

Solving this new equation using the quadratic formula we get two imaginary solutions :
   r =  0.5000 ± 0.8660 i 
Now that we know the value(s) of  r , we can calculate  s  since  s  is  ∛ r  

Since we are speaking 3rd root, each of the two imaginary solutions of has 3 roots

Tiger finds these roots using de Moivre's Formula

The 3rd roots of   0.500 + 0.866 i   are:

  s =  0.940 + 0.342 i 
  s = -0.766 + 0.643 i 
  s = -0.174 -0.985 i 

3rd roots of   0.500- 0.866 i  :
  s = -0.174  + 0.985 i  
  s = -0.766 - 0.643 i 
  s =  0.940 - 0.342 i 
 

10 solutions were found :

1.   s = 0.940 - 0.342 i
2.   s = -0.766 - 0.643 i
3.   s = -0.174 + 0.985 i
4.   s = -0.174 -0.985 i
5.   s = -0.766 + 0.643 i
6.   s = 0.940 + 0.342 i
7.  x  =  0
8.  s  =  0
9.  o  =  0
10.  c  =  0