Solve the equation on the interval [0, 2π). sin²x - cos²x = 0

Solution:

Given, sin²x - cos²x = 0

We have to find the solution on the interval (0, 2π).

We know, cos (2x) = cos²(x) - sin²(x)

So, sin²x - cos²x = -cos(2x)

In general, cos(u) = 0

u = n𝜋/2 for some n𝜖Z

Thus, we have

 sin²x - cos²x = 0

-cos(2x) = 0

2x = n𝜋/2 for some n𝜖Z

x = n𝜋/4 for some n𝜖Z

Restricting values to the interval (0, 2𝜋)

x 𝜖 {𝜋/4, 3𝜋/4, 5𝜋/4, 7𝜋/4}

Therefore, the solution on the interval [0, 2𝜋) are {𝜋/4, 3𝜋/4, 5𝜋/4, 7𝜋/4}.

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Solve the equation on the interval [0, 2π). sin²x - cos²x = 0

Summary:

The solutions to the equation sin²x - cos²x = 0 on the interval [0, 2π) are  {𝜋/4, 3𝜋/4, 5𝜋/4, 7𝜋/4}.