Question

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

Answer 1

`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`