prove that: sin²(3x) + cos²(3x) + sin²(x) - cos²(x) = 0

Answer 1

well, easy
sin^2 (3x)+cos^2(3x)=1
that leaves on the left
1-cos^2 x + sin^2 x
but 1-cos^2 x=sin^2 x
so, now you are left with
sin^2 x + sin^2 x which is NOT zero. The equation is false.

Answer 2

sin²(3x) + cos²(3x) = 1

1 + sin²(x) - cos²(x) = 2sin²(x)
not zero

I suspect you have a typo. A true assertion is

sin²(3x) + cos²(3x) - sin²(x) - cos²(x) = 0

Answer 3

thank you
then how do you solve : sin²(3x) + cos²(3x) - sin²(x) - cos²(x) = 0

Answer 4

come on you know that

sin²(u) + cos²(u) = 1

so, the equation reduces to

sin²(3x) + cos²(3x) - (sin²(x) + cos²(x)) = 0
1 - 1 = 0

Answer 5

thank you steve