Solve the equation giving all values of x which lie between 0 degrees and 360 degrees

Question

1) 2sec x =4.6

2) 3cos^2 x + cos x-2 =0

3) Sech x =0.554

Answer 1

Hi Jack

 

1)

2 sec x = 4.6

sec x = 2.3

Since sec = 1/cos, then cos x = 1/4.6 = .217

arccos (.217) = 77.4 degrees

Cos is positive in the 1st and 4th quadrant so the solution is 77.4 and 282.6 (or 360-77.4)

 

2)

3 cos²x + cos x - 2 = 0

This is a quadratic equation of trig functions, so let's factor

(3 cos x -2)(cos x + 1) = 0

3 cos x -2 = 0   and   cos x + 1 = 0

3 cos x = 2       and    cos x = -1

cos x = 2/3

x = 48.2 and              x = 180

311.8

 

3)

Sech x = 0.554

sech x = 1/cosh x = 2/( e^x + e^-x )

cosh x = 1/.554 = 1.805

arccosh (1.805) = 68.5 or 291.5