<PRE><B>Question 729916</B>
Remember the identity that {{{sin^2(x) + cos^2(x) = 1}}}.  Whenever there is sin or cos squared and the other not squared, it&#39;s a good bet to use that identity.&lt;P&gt;
{{{cos^2(x) = 1 - sin^2(x)}}}&lt;P&gt;
The equation becomes {{{1-sin^2(theta) + 2*sin(theta) = 2}}}&lt;P&gt;
Subtract 2 from both sides.&lt;P&gt;
{{{-sin^2(theta) + 2*sin(theta) - 1 = 0}}}&lt;P&gt;
Multiply each term by negative 1.&lt;P&gt;
{{{sin^2(theta)-2*sin(theta)+1 = 0}}}&lt;P&gt;
Substitute {{{sin(theta)=x}}}&lt;P&gt;
{{{x^2 -2x + 1 = 0}}}&lt;P&gt;
That&#39;s a quadratic polynomial.&lt;P&gt;
{{{(x-1)(x-1) = 0}}} so x=1&lt;P&gt;
{{{x = sin(theta) = 1}}}&lt;P&gt;
Arcsin (1) = 90 degrees.  That&#39;s only one possible answer, because sin repeats every 2*pi radians (360 degrees).  The complete answer is pi + 2*pi*n radians, or 90 + 360n degrees.  
&lt;P&gt;Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)
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