what would be the solutions to (sin2x+cos2x)^2=1?

Question

I know that this can be expanded to (sin2x+cos2x) (sin2x+cos2x)=1, which can be multiplied to be sin^2(4x)+2sin(2x)cos(2x)+cos^2(4x)=1. Or am I missing something here? I'm not sure what to do after this. Please help! 

Answer 1

Hyun,

Your expansion is a little off.  sin(2x) * sin(2x) ≠ sin²(4x), it just equals sin²(2x).

The sine of an expression still represents a single value, so squaring the sine does not square

the expression.  The proper expansion would be:

 

sin²(2x) + 2sin(2x)cos(2x) + cos²(2x) = 1

 

using the trig identity sin²A + cos²A = 1, we end up with:

 

2sin(2x)cos(2x) = 0

 

applying the zero product rule we get:

 

sin(2x) = 0      and      cos(2x) = 0

 

x = 0, π/2, π, 3π/2, 2π, etc.   and   x = π/4, 3π/4, 5π/4, 7π/4, etc.

 

Hope this helps.