solve 5sin2x + 12cos2x =11 for 0°«x«180°

To solve the equation 5sin(2x) + 12cos(2x) = 11 for 0° ≤ x ≤ 180°, we can use trigonometric identities to simplify it.

Let's start by using the double-angle identities for sine and cosine:

sin(2x) = 2sin(x)cos(x)
cos(2x) = cos²(x) - sin²(x)

Replacing sin(2x) and cos(2x) in the equation, we get:

5(2sin(x)cos(x)) + 12(cos²(x) - sin²(x)) = 11

Now, let's simplify further:

10sin(x)cos(x) + 12cos²(x) - 12sin²(x) = 11

Using the Pythagorean identity sin²(x) + cos²(x) = 1, we can replace cos²(x) with 1 - sin²(x):

10sin(x)cos(x) + 12(1 - sin²(x)) - 12sin²(x) = 11

Expanding and rearranging terms:

10sin(x)cos(x) + 12 - 12sin²(x) - 12sin²(x) = 11

Combining like terms:

-24sin²(x) + 10sin(x)cos(x) + 12 = 11

Rearranging the equation to set it equal to zero:

-24sin²(x) + 10sin(x)cos(x) - 1 = 0

We now have a quadratic equation in terms of sin(x). To solve this, we can let sin(x) = t and solve for t using the quadratic formula.

The quadratic formula is given by:

t = (-b ± √(b² - 4ac)) / 2a

In this case, a = -24, b = 10cos(x), and c = -1.

Substituting the values into the quadratic formula:

t = (-(10cos(x)) ± √((10cos(x))² - 4(-24)(-1))) / 2(-24)

Simplifying:

t = (-10cos(x) ± √(100cos²(x) - 96)) / (-48)

To solve for sin(x), we can substitute t back in:

sin(x) = t

Therefore, to solve the equation 5sin(2x) + 12cos(2x) = 11 for 0° ≤ x ≤ 180°, we need to use the quadratic formula to solve for t and then find the corresponding values of x using the inverse sine function (sin⁻¹) or a calculator.