SOLUTION: Solve for x if: 5sin^2x-5cosx=-3 ; x is between 0deg and 360deg

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Question 972493: Solve for x if:
5sin^2x-5cosx=-3 ; x is between 0deg and 360deg

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
your equation is:

5sin^2(x) - 5cos(x) = -3

since sin^2(x) is equal to 1 - cos^2(x), then you equation can be converted to:

5 - 5cos^2(x) - 5cos(x) = -3

multiply both sides of the equation by -1 and you get:

-5 + 5cos^2(x) + 5cos(x) = 3

subtract 3 from both sides of the equation and re-arrange the terms in descending order of exponential degree and you get:

5cos^2(x) + 5cos(x) - 8 = 0

this is a quadratic equation that you can solve using the quadratic equation.

you will get:

cos(x) = .8601470509
cos(x) = -1.860147051

solve for x by taking the arc-cosine of those values and you get:

x = 30.66690224 degrees in the first quadrant.
x = invalid value for arc-cosine (-1.86...) because the maximum and minimum value of cos(x) is between -1 and 1.

so you have one solution.

x = 30.66690224 degrees.

that's in the first quadrant.

cosine is also positive in the fourth quadrant.

value of x in the fourth quadrant is 360 - 30.66690224 degrees = 329.3330978.

in the interval between 0 degrees and 360 degrees you have two solutions for x that will satisfy the equation.

they are:

x = 30.66690224 degrees
x = 329.3330978 degrees

here's a graph of the equations of:

y = 5sin^2(x) - 5cos(x)
y = -3

the intersections are where both equations have the same value of y for the same values of x.

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