4cos^2x-5sinxcosx-6=0
find the x :( please
since cos2x = 2cos^2 x - 1, we have
2cos2x - 5/2 sin2x - 4 = 0
or
4cos2x - 5sin2x - 8 = 0
4cos2x = 5sin2x + 8
16cos^2 2x = 25sin^2 2x + 80sin2x + 64
16 - 16sin^2 2x = 25sin^2 2x + 80sin2x + 64
41sin^2 2x + 80sin2x + 48 = 0
now just solve for sin2x using the quadratic formula.
To find the value of x in the given equation, we can use the quadratic formula since the equation is quadratic in terms of cos(x). The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the given equation with the quadratic form, we have:
a = 4
b = -5sin(x)
c = -6
Substituting these values into the quadratic formula, we get:
cos(x) = [5sin(x) ± √(25sin^2(x) - 4(4)(-6))] / (2(4))
Simplifying further:
cos(x) = [5sin(x) ± √(25sin^2(x) + 96)] / 8
To solve for x, we need to find the values of sin(x) that satisfy this equation. However, since sin(x) is not given, we cannot directly solve for x using the given equation. Additional information is needed to determine the value of sin(x) or to simplify the equation further.