Question 1045455
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Please help. I would like the answer in radians. Thanks!
2cos^2x + sinx -1 =0
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<pre>
First what we need to do is to reduce the given equation to the quadratic equation for sin(x).
For it, use the identity cos^2(x) = 1-sin^2(x) and replace cos^2(x) in the equation by this expression. You will get

2*(1-sin^2(x) + sin(x) - 1 = 0,   or

2 - 2sin^2(x) + sin(x) - 1 = 0,   or

2sin^2(x) - sin(x) - 1 = 0.

Factor left side:

(2sin(x) + 1)*(sin(x) - 1) = 0.

Now the equation deploys in two independent equations:


1.  2sin(x) + 1 = 0  --->  sin(x) = {{{-1/2}}}  --->  x = {{{7pi/6}}}  and/or  x = {{{11pi/6}}}.


2.  sin(x) - 1 = 0  --->  sin(x) = 1  --->  x = {{{pi/2}}}.


<U>Answer</U>.  The solutions are  x = {{{pi/2}}},  {{{7pi/6}}} and  {{{11pi/6}}}.
</pre>

<TABLE> 
  <TR>
  <TD> 

{{{graph( 330, 330, -0.5, 6.5, -2.5, 2.5,
          2*(cos(x))^2 + sin(x) -1
)}}}


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

  </TD>
  </TR>
</TABLE>