SOLUTION: Solve cos(2x)+sin(x)=1in[0,2pi)

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Question 868732: Solve cos(2x)+sin(x)=1in[0,2pi)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Use the trigonometric identity cos%282x%29=%28cos%28x%29%29%5E2-%28sin%28x%29%29%5E2 .
Since %28cos%28x%29%29%5E2=1-%28sin%28x%29%29%5E2 , it can be re-arranged to
cos%282x%29=1-2%28sin%28x%29%29%5E2 .
Substituting into +cos%282x%29%2Bsin%28x%29=1 you get
1-2%28sin%28x%29%29%5E2%2Bsin%28x%29=1<--->-2%28sin%28x%29%29%5E2%2Bsin%28x%29=0 (subtracting 1 from both sides of the equal sign)
That is a quadratic equation in sin%28x%29 .
It looks complicated, but if you rename y=sin%28x%29 ,
you can re-write it and it looks very simple:
-2y%5E2%2By=0<-->2y%5E2-y-0<-->y%282y-1%29=0-->system%28y=0%2C%22or%22%2Cy=1%2F2%29
So, -2%28sin%28x%29%29%5E2%2Bsin%28x%29=0 is true when
system%28sin%28x%29=0%2C%22or%22%2Csin%28x%29=1%2F2%29 .
In %22%5B+0+%2C%222pi%22%29%22 ,
sin%28x%29=0 --> system%28highlight%28x=0%29%2C%22or%22%2Chighlight%28x=pi%29%29 and
sin%28x%29=1%2F2 --> system%28highlight%28x=pi%2F6%29%2C%22or%22%2Chighlight%28x=5pi%2F6%29%29