cos(pi/2(x+1))=cospi/2 x
cos((pi/2)x + pi/2) = cos(pi/2)x
cos((pi/2)x)cos(pi/2) - sin((pi/2)x)sin(pi/2) = cos (pi/2)x
0 - sin((pi/2)x = cos (pi/2)x
divide both sides by -cos(pi/2)x
tan (pi/2)x = -1
we know tan (pi/2) = +1 , so (pi/2)x must in the II or IV quadrant.
so (pi/2)x = pi - pi/2 or (pi/2)x = 2pi - pi/2
x = 1/2 or x = 3/2
but the period of tan (pi/2)x = 2
(unlike the sine or cosine which would indeed by 4, but I was solving tan at the end)
so the general solution is
1/2 + 2n but also 3/2 + 2n
the book missed that one, but when you sub in 3/2 it works in the original equation, and the general solution of
1/2 + 2n does not produce 3/2
Solve each equation , then write the general solution
cos(pi/2(x+1))=cospi/2 x
x=1.5, 3.5
general solution: 1/2+4n
but the back of the book says that the general solution is 1/2+2n
isn't the period 4?
3 answers
hmm.. how did you get
cos((pi/2)x)cos(pi/2) - sin((pi/2)x)sin(pi/2) = cos (pi/2)x
also, do you have to solve for tan?
cos((pi/2)x)cos(pi/2) - sin((pi/2)x)sin(pi/2) = cos (pi/2)x
also, do you have to solve for tan?
I just learnt to find the intersections of these 2 equations on a graphing calc. and then write the general solution based on the intersections.