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
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
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.
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