solve on the interval 0 < equal to theta < 2pi
D) cos(2theta) = 2 - 2 sin^2 theta
E) cos (2theta) + 5 cos theta + 3 = 0
please help and explain, thanks!!
2007-10-07 09:52:00 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Hey! You were supposed to learn how to do these from the solution someone gave you for C) Follow their instruction and set 2 theta = 2x to make the notation easier. This gives
cos2x = 2-2sin2^x. What was the next hint you were given on C)? Express everything in terms of the same function! Check your list of identities for cos 2x and wirte it with identities for just sinx or sin^2x. Then simplify and try to factor. If stuff cancels out and you get a false statement like 0=1 you know there is no solution. If you get something like sinx = 5 there is no solution for that factor since sinx is always between -1 and +1. Now work on your own homework. Struggle and don't give up and you will learn something.
2007-10-07 10:09:47 · answer #1 · answered by baja_tom 4 · 0⤊ 0⤋
D)
It has no solution.
E)
cos(2x) + 5 cos(x) + 3 = 0
2cos^2 (x) - 1 + 5 cos(x) + 3 = 0
2 cos^2(x) + 5 cos(x) + 2 = 0
2cos^2 (x) + 4 cosx + cosx + 2 = 0
2 cosx(cosx + 2) + 1(cosx + 2) = 0
(cosx + 2)(2cosx - 1) = 0
2 cosx = 1
cosx = 1/2
x = pi/3, 4pi/3,
2007-10-07 10:10:48 · answer #2 · answered by mohanrao d 7 · 0⤊ 0⤋
Already did this. Check my previous answer.
2007-10-07 09:56:49 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋