Question 729916
{{{cos^2(theta) + 2sin(theta) = 2}}}
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Since {{{sin^2(theta) + cos^2(theta) = 1}}} (a fundamental trigonometric identity), you can change this to {{{cos^2(theta) = 1 - sin^2(theta)}}} using basic algebra subtraction on both sides and substitute this into your equation for {{{cos^2(theta)}}}.
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{{{1 - sin^2(theta) + 2sin(theta) = 2}}}
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Move everything to the right side: {{{0 = sin^2(theta) - 2sin(theta) + 1}}}
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Factor: {{{0 = (sin(theta) - 1)(sin(theta) - 1)}}}
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Solve for {{{sin(theta)}}}: {{{sin(theta) - 1 = 0}}} ==> {{{sin(theta) = 1}}}
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Now you have to think: for what angles is the sine equal to 1? Well, for {{{pi/2}}}. However, it is also 1 at every complete revolution (every {{{2pi}}} from {{{pi/2}}} -- that is, {{{pi/2 + 2pi*n}}}, where "n" is an integer {think {{{-3pi/2}}}, {{{5pi/2}}}, etc.}). Therefore, the solution to the equation is {{{theta = pi/2 + 2pi*n}}}.
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IF you have a domain restriction in the problem, only include solutions that are within the domain. For example, if you have a restriction of {{{0 <= theta < 2pi}}}, then your solution would be {{{theta = pi/2}}} only because that is the only solution in {{{pi/2 + 2pi*n}}} that is between 0 and {{{2pi}}}.
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I hope this helped! Good luck!