Find x-coordinates where the curve y=f(x)=a+bcos(2x) intersects the x-axis
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What is the value of f(x) when it intersects the x-axis?I have worked out the values of a and b, but i cant seem to find the x-coordinates
The function f is defined by f(x)=a+bcos(2x), for ( 0 /= x /= pi. It is given that F(0)=-1 and f(1/2 pi)=7.
I've worked out a=3 and b=-4
Please help
Steven G
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What is F? You defined f, but not FI have worked out the values of a and b, but i cant seem to find the x-coordinates
The function f is defined by f(x)=a+bcos(2x), for ( 0 /= x /= pi. It is given that F(0)=-1 and f(1/2 pi)=7.
I've worked out a=3 and b=-4
Please help
D
Deleted member 4993
Guest
Then - to find x-coordinates of of x- intersect solve for 'x' in:
0=a+bcos(2x), for ( 0 /= x /= pi.)
0=3-4cos(2x)
3/4=cos(2x)
1/2(arccos(3/4))=x
0.361=x
Ive only got 1 answer, did I miss a step? As t meets the x-axis in 2 places
Steven G
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So the cos(2x) is positive, after all 3/4 >0. Which quadrants have positive values for cosine?? Try to proceed with that hint. If you still are having trouble, then at least answer my question and someone will help you.0=3-4cos(2x)
3/4=cos(2x)
1/2(arccos(3/4))=x
0.361=x
Ive only got 1 answer, did I miss a step? As t meets the x-axis in 2 places
Steven G
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Sorry, but no, this is not correct. Please tell us two things now. Which quadrant or quadrants, if any, have cos(x) positive. Also let us know which quadrant pi-0.361 is in.pi-0.361=2.78?
x=2.78
Is this correct?
Thank you so much for the help
Cos(x) is positive in the 1st and 4th quadrant
pi-0.361 is in the 2nd quadrant
I've looked this up it says in the 1st quadrant cos is
- cos(360+theta)=cos(theta)
- cos(theta)=cos(-theta)
In the 4th:
- cos(360-theta)=cos(theta)
Doesnt it have to be in the 2nd quadrant because its o<= x <= pi?