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What is cos5x = 0 between interval 0<180<360 degreesI tried doing it by making...
What is cos5x = 0 between interval 0<180<360 degrees
I tried doing it by making cos5x =
cos(3x+2x) = 0, then i follow the trig equiation cosAcosB - sinAsinB
cos3xcos2x - sin3xsin2x
from here i get: 2cos^2x - 1 . cosx - 2sinx . cosx. sinx
then.. 2cos^3x - cosx - 2sin^2x . cosx
cosx (2cos^2x - 1 - 2sin^2x) = 0
cosx = 90 here.. (one of the answer)
2cos^2x - 1 - 2 sin^2x will now become
2 - 2sin^2x - 1 - 2sin^2x , due to the identity formula
from here i got: 4sin^2x = 1
sin^2x = 1/4
Sinx = the root of 1/4 ( + or - ) of course
For + i got 30, 150
for - i got 210, 330.
Is this correct please help :( im confuse if this method actualy works or not, because another method which involves inverse cos 5x will give various answers which are different, and i did cos3x before and my answer for cos 3x is similar to the cos 5x, but then when i did inverse method for cos3x it turns out fine. helppp, i got 10 values for x when i use the inverse method on cos5x. Im confused ;(
1 Answer | add yours
High School Teacher
Given that cos5x = 0.
We know from trigonometric identities that cos(x) = 0 when x= 30, 150, 210, and 330.
==> cos5x = 0 when (5x)= 30, 150, 210, and 330
To find the value of x, we will divide by 5.
==> x = 30/5 , 150/5, 210/5, and 330/5
==> x = 6, 30, 42, and 66 degrees.
Posted by hala718 on August 12, 2011 at 6:58 PM (Answer #1)
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