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.