Question

6sin^2x=5sin(3pi/2-x)+2

Answer 1

Решение смотрите в приложении...

Answer 2

To solve the equation 6sin^2x = 5sin(3π/2 - x) + 2, we'll start by simplifying the right side:

Using the angle difference identity sin(3π/2 - x) = sin(3π/2)cos(x) - cos(3π/2)sin(x) = -cos(x)

Now, the equation becomes:

6sin^2x = 5(-cos(x)) + 2

Let's rearrange the equation:

6sin^2x = -5cos(x) + 2

Now, we'll apply the Pythagorean identity sin^2(x) + cos^2(x) = 1 to express sin^2(x) in terms of cos(x):

6(1 - cos^2(x)) = -5cos(x) + 2

Expand and rearrange the equation:

6 - 6cos^2(x) = -5cos(x) + 2

Rearrange again:

6cos^2(x) - 5cos(x) + 4 = 0

Now, we have a quadratic equation in terms of cos(x). We can solve it by factoring or using the quadratic formula. However, this equation doesn't factor nicely, so let's use the quadratic formula:

cos(x) = [-(-5) ± √((-5)^2 - 4(6)(4))] / (2(6))

Simplifying:

cos(x) = [5 ± √(25 - 96)] / 12

cos(x) = [5 ± √(-71)] / 12

Since we have a negative value inside the square root, there are no real solutions for cos(x) in this equation. Therefore, the original equation 6sin^2x = 5sin(3π/2 - x) + 2 does not have any real solutions.