How do you solve this trig identity problem without factoring?

I just used @ to represent theta:

sin^4@ + 2sin^2@cos^2@ + cos^4@ = 1

sin^4 + 2sin^2 cos^2 + cos^4 = (sin^2 + cos^2)^2

1^2 = 1

Didn't you just factor that? (perfect square trinomial)

I'm looking for a way to solve it without factoring..

To solve this trig identity problem without factoring, we can use the Pythagorean identity: sin^2@ + cos^2@ = 1. Let's break down the steps:

1. Start with the given trigonometric expression:
sin^4@ + 2sin^2@cos^2@ + cos^4@ = 1

2. Use the Pythagorean identity to replace sin^2@ and cos^2@:
(sin^2@ + cos^2@)^2 - 2sin^2@cos^2@ + cos^4@ = 1

3. Simplify the expression:
1^2 - 2sin^2@cos^2@ + cos^4@ = 1

4. Combine like terms on the left side:
1 - 2sin^2@cos^2@ + cos^4@ = 1

5. Subtract 1 from both sides to isolate the trinomial expression:
- 2sin^2@cos^2@ + cos^4@ = 0

6. Factor out common terms from the trinomial expression:
cos^2@(-2sin^2@ + cos^2@) = 0

7. Apply the zero product property:
cos^2@ = 0 or -2sin^2@ + cos^2@ = 0

8. Solve for cos^2@ = 0:
cos^2@ = 0 implies that @ = 90 degrees or @ = 270 degrees (plus/minus multiples of 180 degrees)

9. Solve for -2sin^2@ + cos^2@ = 0:
-2sin^2@ + cos^2@ = 0
Rearrange the terms:
cos^2@ = 2sin^2@
Divide by cos^2@ (since cos^2@ cannot be zero):
1 = 2tan^2@

10. Solve for tan^2@:
Divide by 2:
1/2 = tan^2@

11. Take the square root of both sides to solve for tan@:
√(1/2) = tan@

Therefore, the solution for this trig identity problem without factoring is @ = 45 degrees or @ = 225 degrees (plus/minus multiples of 180 degrees).