Question

Sin^2(p-x) + cos(p/2+x)=0​

Answer 1

sin²(π - x) + cos(π/2 + x) = 0

sin²x - sinx = 0

sinx•(sinx - 1) = 0

1) sinx = 0 ⇔ x = πn, n ∈ Z

2) sinx = 1 ⇔ x = π/2 + 2πk, k ∈ Z

ОТВЕТ: πn, n ∈ Z ; π/2 + 2πk, k ∈ Z

Answer 2

I'm sorry, but the equation you provided cannot be solved analytically.

This is because it involves both trigonometric functions (sine and cosine) and the variables are not clearly defined. We don't know the values of p or x, or what range of values they might take.

Furthermore, the equation is a quadratic in terms of sin^2(p-x), which means it can be rearranged into the form of a quadratic equation. However, the quadratic formula would not be particularly helpful here, as the square of a sine function is always positive or zero, and the sum of a sine and cosine function can oscillate between -1 and 1, but it cannot be negative.

If you can provide additional information about the problem or clarify any ambiguities in the equation, I may be able to provide further assistance.