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The function y=sin^3x has five stationary points in, negative pi is less than or equal to x and positive pi is greater than or equal to x.
Find dy/dx and find the coordinates of the five stationary points:
I have found dy/dx: 3cosxsin^2x and made it equal to zero but don't know where to go from there!! Any help would be greatly appreciated.
Find dy/dx and find the coordinates of the five stationary points:
I have found dy/dx: 3cosxsin^2x and made it equal to zero but don't know where to go from there!! Any help would be greatly appreciated.
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#2
The equation 3cosxsin^2x=0 has solution between -pi and pi where either cos x is zero or sin^2x is zero.
sin^2 x is zero at -pi, 0 and pi. Cos x is zero at -pi/2 and +pi/2 and so you have turning points at -pi, -pi/2, 0, pi/2 and pi.
To see them, plot sin^3x and 3cosxsin^2x on the same axes. (fooplot.com is good). (attached)
![Name: save.png
Views: 89
Size: 19.8 KB]()
sin^2 x is zero at -pi, 0 and pi. Cos x is zero at -pi/2 and +pi/2 and so you have turning points at -pi, -pi/2, 0, pi/2 and pi.
To see them, plot sin^3x and 3cosxsin^2x on the same axes. (fooplot.com is good). (attached)
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#3
(Original post by metaljoe)
The function y=sin^3x has five stationary points in, negative pi is less than or equal to x and positive pi is greater than or equal to x.
Find dy/dx and find the coordinates of the five stationary points:
I have found dy/dx: 3cosxsin^2x and made it equal to zero but don't know where to go from there!! Any help would be greatly appreciated.
The function y=sin^3x has five stationary points in, negative pi is less than or equal to x and positive pi is greater than or equal to x.
Find dy/dx and find the coordinates of the five stationary points:
I have found dy/dx: 3cosxsin^2x and made it equal to zero but don't know where to go from there!! Any help would be greatly appreciated.
Use one of those identities to replace the sin^2x and rewrite in terms of just cosx
Factorise.
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(Original post by nerak99)
The equation 3cosxsin^2x=0 has solution between -pi and pi where either cos x is zero or sin^2x is zero.
sin^2 x is zero at -pi, 0 and pi. Cos x is zero at -pi/2 and +pi/2 and so you have turning points at -pi, -pi/2, 0, pi/2 and pi.
To see them, plot sin^3x and 3cosxsin^2x on the same axes. (fooplot.com is good). (attached)
![Name: save.png
Views: 89
Size: 19.8 KB]()
The equation 3cosxsin^2x=0 has solution between -pi and pi where either cos x is zero or sin^2x is zero.
sin^2 x is zero at -pi, 0 and pi. Cos x is zero at -pi/2 and +pi/2 and so you have turning points at -pi, -pi/2, 0, pi/2 and pi.
To see them, plot sin^3x and 3cosxsin^2x on the same axes. (fooplot.com is good). (attached)
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(Original post by gdunne42)
When solving trigonometric equations in C2 you used identities to simplify equations that had a mixture of trig functions.
Use one of those identities to replace the sin^2x and rewrite in terms of just cosx
Factorise.
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When solving trigonometric equations in C2 you used identities to simplify equations that had a mixture of trig functions.
Use one of those identities to replace the sin^2x and rewrite in terms of just cosx
Factorise.
Posted from TSR Mobile
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#6
(Original post by metaljoe)
I will use the advice. Thank you !!
I will use the advice. Thank you !!
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#7
(Original post by nerak99)
It is worth pointing out that 3cosxsin^2x=0 is already factorised and going through the process described by gdunne42 is possible but entirely redundant.
It is worth pointing out that 3cosxsin^2x=0 is already factorised and going through the process described by gdunne42 is possible but entirely redundant.
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