SOLUTION: find x where 0<\x<\360 A) sec x + 4cos x =5 Plz help me i try lots but invain. thx in adv

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Question 1116912: find x where 0<\x<\360
A) sec x + 4cos x =5
Plz help me i try lots but invain.
thx in adv
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
your equation is sec(x) + 4cos(x) = 5

sec(x) is equivalent to 1/cos(x).

your equation becomes 1/cos(x) + 4cos(x) = 5

multiply both sides of the equation by cos(x) to get:

1 + 4cos^2(x) = 5cos(x)

subtract 5cos(x) from both sides of the equation to get 1 + 4cos^2(x) - 5cos(x) = 0

reorder the terms in descending order of degree to get:

4cos^2(x) - 5cos(x) + 1 = 0

factor this quadratic equation to get:

cos(x) = .25 or cos(x) = 1

in the first quadrant, cos(x) = 1 when x = 0 degrees.

in the first quadrant, cos(x) = .25 when x = 75.52248781.

cos(x) is positive in the first and fourth quadrant.

in the fourth quadrant, 0 degrees is equal to 360 - 0 = 360 degrees.

in the fourth quadrant, 75.52248781 degrees is equal to 360 - 75.52248781 = 284.4775122 degrees.

your solution is that:

x = 0, 75.52248781, 284.4775122, 360 degrees in the interval from 0 to 360 degrees.

that would be the interval 0 <= x <= 360 degrees.

in the interval 0 < x < 360 degrees, then your only solutions are x = 75.52248781 or x = 284.4775122.

here's a graph of your equation with the possible solutions shown.