Question 608023: Hi, I hope someone helps me with this trigonometric equations problem.
How do you get the solution set of cos^2 2x + 3sin2x = 3 ?
Also please include the steps. I don't understand our lesson well, especially the part where the reference angle enters. I don't get it at all. Thank yoy for your help!
Found 2 solutions by Alan3354, htmentor:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! cos^2 2x + 3sin2x = 3
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cos^2 = 1 - sin^2
1 - sin^2 + 3sin = 2
sin^2 - 3sin + 1 = 0
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=5 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 2.61803398874989, 0.381966011250105.
Here's your graph:
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sin(2x) =~ 0.381966 (Ignore the sin > 1 unless you want complex angles)
2x =~ 0.39192 + 2n*pi radians, n = 0,1,2,3...
x = 0.19596 radians + n*pi, n = 0,1,2,3...
x = pi - 0.19596 + n*pi, n = 0,1,2,3...
Answer by htmentor(1343)  (Show Source):
You can put this solution on YOUR website! cos^2 2x + 3sin2x = 3
First, we can use the identity cos^2(2x) = 1 - sin^2(2x) to obtain a quadratic in sin(2x):
1 - sin^2(2x) + 3sin(2x) - 3 = 0
sin^2(2x) - 3sin(2x) + 2 = 0
This can be factored as:
(sin(2x) - 1)(sin(2x) - 2) = 0
The LHS will be equal to 0 if either sin(2x) = 1, or sin(2x) = 2
Since the sine of any angle can never be >1, the only solutions are obtained for sin(2x) = 1
In the 1st quadrant, this gives 2x = 90 deg -> x = 45 deg [ /4]
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