How do you solve cos(2x)+cos(2x)*sin(2x)=0 in the interval [0, 2pi]?

1 Answer
Feb 7, 2016

x=pi/2,pi/4, 3pi/4

Explanation:

let f(x)= cos(2x) + cos(2x)sin(2x)
Solve for x in f(x)= 0

cos(2x) + cos(2x)sin(2x) = 0
cos(2x)[1+sin(2x)] = 0
in there equation , the only way f(x) wil be 0
if the multiplier of the multiplicand is 0

so cos(2x)=0
for cos(2x) to be 0
2x=pi/2, 3pi/2
x = pi/4, 3pi/4

Or 1+sin(2x)=0
sin(2x)=-1
2x= pi
x=pi/2