SOLUTION: Solve: cos^2(theta)+ 2sin(theta)=2 Please help. I have no Idea what to do.

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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!

.
Since (a fundamental trigonometric identity), you can change this to using basic algebra subtraction on both sides and substitute this into your equation for .
.

.
Move everything to the right side:
.
Factor:
.
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 .
.
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 .
*
I hope this helped! Good luck!

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