Find all real numbers in the interval [0,2pi) that satisfy the equation.?

Question

5 cos^2 x+9 cos x= -4

round to the nearest tenth

Answer 1

Let y = cos x

Then 5 cos^2 x+9 cos x= -4 --> 5 cos^2 x+9 cosx+4=0 becomes 5y^2+9y+4=0 or (5y + 4)(y+1)=0

Hence y = -1 or y =-4/5. In other words, cosx=-1 or cosx=-4/5. So x = 3.1, x = 2.5, or x = 3.8 are approximate solutions.

Hope this helps

By the way, there must be 2 distinct solutions for cos(-4/5), which are approximated above

Answer 2

This is a slight variation on the classic "quadratic roots" problem. For the sake of clarity, I'll make the, perhaps "obvious" simplification of

u = cos x;

which, with a little rearranging gives

5u² + 9u + 4 = 0.

This can be easily factored into a product of two linear binomials:

(5u + 4)(u + 1) = 0;

which, assuming one or the other term equals zero, has the solutions:

u = -4/5 and u = -1.

Now since u = cos x, that makes

cos x = -4/5, and cos x = -1.

If you are familiar with trigonometry, cos x = -1 has the well known solution of

x = π.

The solution to cos -4/5 is not easy to calculate on paper, but it does correspond to the well known "3,4,5 triangle." In any case the solution is not a rational multiple of π, the best way to approximate the value is to simply use the inverse cosine function on a calculator:

x = acos -4/5.

Hope that makes sense,
~W.O.M.B.A.T.

Answer 3

This can be rearranged into
[(5cos x)+4]*[(cos x)+1]=0
so you need to solve for
cos x=-1
and
cos x=-4/5

Answer 4

15sin(x)^2 = 4sin(x) + 3 15sin(x)^2 - 4sin(x) - 3 = 0 sin(x) = (4 +/- sqrt(sixteen + one hundred eighty)) / 30 sin(x) = (4 +/- sqrt(196)) / 30 sin(x) = (4 +/- 14) / 30 sin(x) = (2 +/- 7) / 15 sin(x) = 9/15 , -5/15 sin(x) = 3/5 , -a million/3 x = arcsin(3/5) , arcsin(-a million/3)