Question 729916: Solve: cos^2(theta)+ 2sin(theta)=2
Please help. I have no Idea what to do.
Found 3 solutions by stanbon, fcabanski, Leaf W.:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! cos^2(theta)+ 2sin(theta)=2
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You have to know cos^2 + sin^2 = 1
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Your Problem:
(1-sin^2) + 2sin = 2
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sin^2 - 2sin + 1 = 0
Factor:
(sin-1)^2 = 0
sin(theta) = 1
theta = pi/2
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Cheers,
Stan H.
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Answer by fcabanski(1391) (Show Source):
You can put this solution on YOUR website! Remember the identity that  . Whenever there is sin or cos squared and the other not squared, it's a good bet to use that identity.
The equation becomes 
Subtract 2 from both sides.
Multiply each term by negative 1.
Substitute 
That's a quadratic polynomial.
 so x=1
Arcsin (1) = 90 degrees. That'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.
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.)
Answer by Leaf W.(135)  (Show Source):
You can put this solution on YOUR website! 
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Since  (a fundamental trigonometric identity), you can change this to  using basic algebra subtraction on both sides and substitute this into your equation for  .
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Move everything to the right side: 
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Factor: 
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Solve for  :  ==> 
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Now you have to think: for what angles is the sine equal to 1? Well, for  . However, it is also 1 at every complete revolution (every  from -- that is,  , where "n" is an integer {think  ,  , etc.}). Therefore, the solution to the equation is  .
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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  , then your solution would be  only because that is the only solution in that is between 0 and .
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I hope this helped! Good luck!
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