wowihavefun
07.01.2020 •
Mathematics
Find all solutions in the interval [0,2pi).
cos 2x + sqrt(2) sinx=1
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Ответ:
Note that: cos 2x = cos^2x - sin^2x = (1-sin^2x) - sin^2x = 1 - 2sin^2x.
So, when alternatively written, you have the following equation:
- 2sin^2x + sqrt(2)sinx + 1 = 1
- 2sin^2x + sqrt(2)sinx = 0
Then, let z=sin(x). So you get,
- 2z^2 + sqrt(2)z = 0
z(- 2z + sqrt(2)) = 0
Either z=0, or - 2z + sqrt(2) = 0 ---> z=sqrt(2)/2.
Then, since z=0 or z=sqrt(2)/2, therefore sin(x)=0, or sin(x)=sqrt(2)/2.
Then, for you remains just to list the angles. (Let me know if this is not fair or if you got questions.)
Ответ:
The correct Answer is:
1. Row 3
2. Row 2
6 and 10
Step-by-step explanation:
I Got I right on the assigment Hope I helped :)