Answer – The solutions to the equation (sin 2x + cos 2x)² = 1 are x = (nℼ)/4 with x = 0, ℼ/2, ℼ, (3ℼ)/2, 2ℼ… for even integers (n = 0, 2, 4, 6, 8…) and x = ℼ/4, (3ℼ)/4, (5ℼ)/4, (7ℼ)/4, (9ℼ)/4… for odd integers (n = 1, 3, 5, 7, 9…).

Explanation:

Let’s follow a step-by-step approach to solve the given equation.

First, we expand the equation:

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

Next, we multiply the terms on the left side:

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

We add the common terms:

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

Then, we rearrange the obtained equation:

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

Now, using the trigonometric identity  sin²θ + cos²θ = 1 in the above equation, we get:

1 + 2 sin 2x cos 2x = 1

On simplifying further:

2 sin 2x cos 2x = 0

sin 2x cos 2x = 0

Therefore:

sin 2x = 0 – Equation 1

cos 2x = 0 – Equation 2

Also, from the trigonometric values of special angles:

sin 0 = 0 – Equation 3

cos ℼ/2 = 0 – Equation 4

From Equations 1 and 3, we get:

2x = 0

x = 0

Further, from Equations 2 and 4, we get:

2x = ℼ/2

x = ℼ/4

Hence, x = (nℼ)/4, where n = 0, 1, 2, 3, 4…

Thus, for even integers (n = 0, 2, 4, 6, 8…), the solutions to the given equation are  x = 0,ℼ/2,ℼ,(3ℼ)/2,2ℼ…

And for odd integers (n = 1, 3, 5, 7, 9…), the solutions to the given equation are x = ℼ/4,3ℼ/4,5ℼ/4,7ℼ/4,9ℼ/4…

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