Question 934303
<pre>
{{{sin(2x)+cos(x)=1}}}

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

Substitute {{{sqrt(1-cos^2(x))}}} for {{{sin(x)}}}

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

{{{2sqrt(1-cos^2(x))cos(x)=1-cos(x)}}}

Square both sides:

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

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

{{{-4cos^4(x)+3cos^2(x)+2cos(x)-1 = 0}}}

{{{4cos^4(x)-3cos^2(x)-2cos(x)+1 = 0}}}

1 | 4  0  -3  -2  1
  |<u>    4   4   1 -1</u>  
    4  4   1  -1  0

So we have factored the polynomial in cos(x) as

{{{(cos(x)-1^"")(4cos^3(x)+4cos^2(x)+cos(x)-1^"") = 0}}}

{{{cos(x)-1=0}}},  {{{4cos^3(x)+4cos^2(x)+cos(x)-1 = 0}}}
{{{cos(x)=1}}}
{{{x=2pi*n}}}, n any integer.  That's one solution

{{{4cos^3(x)+4cos^2(x)+cos(x)-1 = 0}}}

That has no rational roots.  Solving it using a TI-84
calculator, we get

{{{cos(x)=0.3478103848}}}
{{{x=1.215561666+2pi*n}}}

But since we had to use a calculator to get the second
answer, we may as well have solved the whole thing by
using the graphing calculator in the first place.

Edwin</pre>