Consider: y=sin(x+pi/9)

(a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

(b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

the "<" are less than or equal to in both cases.

Thank you

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- Jun 4th 2010, 12:53 AMmamannneed description on how to do this
Consider: y=sin(x+pi/9)

(a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

(b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

the "<" are less than or equal to in both cases.

Thank you - Jun 4th 2010, 03:37 AMMush
For part a), when an equation crosses the x axis, the y coordinate is zero, hence the equation becomes:

Now, when you take the inverse sine of 0 you get an infinite number of solutions, but only one of them lies in the interval . Find that solution, then it's just a case of solving algebraically for x.

For part b it is the exact same process except that you need to find take the inverse sine of and find the solution to it which lies in the given interval. - Jun 4th 2010, 04:01 AMmamann
yea but how do i solve for x?

Apparently you cant just apply sin to x and pi/9.

you have to do something where sin(a+b) = (sin a)(cos b) + (sin b)(cos a)

and that gives me:

(sin x)(cos pi/9) + (sin pi/9)(cos x) = 0

Note: pi/9 is in radians which turns out to be 20 degrees.

and i got to here but i dont know what to do after this:

0.94sinx + 0.34cosx = 0 - Jun 4th 2010, 04:11 AMMush
- Jun 4th 2010, 05:47 AMskeeter