# What is cos5x = 0 between interval 0<180<360 degreesI tried doing it by making cos5x = cos(3x+2x) = 0, then i follow the trig equiation cosAcosB - sinAsinB cos3xcos2x - sin3xsin2x from here i...

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 ;(

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.