Question 1209804
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If sinx + sin²x + sin³x = 1,
find cos⁶x - 4cos⁴x + 8cos²x
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        What is written in the post by @CPhill,  it is difficult to call  " a solution ".


        It is just utter gibberish.


        See my solution below.



<pre>
If  {{{sin(x)}}} + {{{sin^2(x)}}} + {{{sin^3(x)}}} = 1,  then

    {{{sin(x)}}} + {{{sin^3(x)}}} = {{{1 - sin^2(x)}}},

    {{{sin(x)*(1+sin^2(x))}}} = {{{cos^2(x)}}}

    {{{sin(x)*(1+(1-cos^2(x)))}}} = {{{cos^2(x)}}}

    {{{sin(x)*(2-cos^2(x))}}} = {{{cos^2(x)}}}.


Square both sides

    {{{sin^2(x)*(2-cos^2(x))^2}}} = {{{cos^4(x)}}}

    {{{(1-cos^2(x))*(4 - 4*cos^2(x) + cos^4(x))}}} = {{{cos^4(x)}}}

    {{{4 - 4cos^2(x) + cos^4(x) - 4cos^2(x) + 4cos^4(x) - cos^6(x)}}} = {{{cos^4(x)}}}


Cancel  {{{cos^4(x)}}}  in both sides;  then combine like terms.  You will get

    {{{4 - cos^6(x) + 4cos^4(x) - 8cos^2(x)}}} = 0,

    {{{cos^6(x) - 4cos^4(x) + 8cos^2(x)}}} = 4.


<U>ANSWER</U>.  {{{cos^6(x) - 4cos^4(x) + 8cos^2(x)}}} = 4.
</pre>

Solved.