How do you solve #sin^2x cos^5x = (sin^2x-2sin^4x+sin^6x) cosx#?
1 Answer
Jun 3, 2016
x is arbitrary.
Explanation:
The RHS
LHS
So, the equation is true for any arbitrary x.
x is arbitrary.
The RHS
LHS
So, the equation is true for any arbitrary x.