Sin^2(x)=5cos(5П/2-x)

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вопрос

To solve the given equation, we can use trigonometric identities to rewrite it.

First, we know that sin^2(x) = 1 - cos^2(x) by the Pythagorean identity.

So, substituting sin^2(x) = 1 - cos^2(x) into the equation, we get:

1 - cos^2(x) = 5cos(5π/2 - x)

Expanding the right side using the cosine difference formula (cos(a-b) = cos(a)cos(b) + sin(a)sin(b)), we get:

1 - cos^2(x) = 5cos(5π/2)cos(x) + 5sin(5π/2)sin(x)
1 - cos^2(x) = 0 - 5cos(x)

Rearranging and simplifying the equation, we get:

cos^2(x) - 5cos(x) + 1 = 0

Now we have a quadratic equation in terms of cos(x). We can solve this equation by using the quadratic formula:

cos(x) = [5 ± sqrt(5^2 - 411)] / 2

cos(x) = [5 ± sqrt(25 - 4)] / 2

cos(x) = [5 ± sqrt(21)] / 2

Therefore, the solutions for cos(x) are:

cos(x) = (5 + sqrt(21)) / 2 or cos(x) = (5 - sqrt(21)) / 2

Since cos(x) = sin(π/2 - x), we can find the values of x using the inverse sin function:

x = π/2 - sin^(-1)((5 + sqrt(21)) / 2) or x = π/2 - sin^(-1)((5 - sqrt(21)) / 2)

These are the solutions for the given trigonometric equation.