SOLUTION: solve for solutions over the interval (0 deg, 360 deg)
4 cos^2 theta + 4 cos theta = 1
i believe the quadratic formula applies but am stuck.
Algebra ->
Trigonometry-basics
-> SOLUTION: solve for solutions over the interval (0 deg, 360 deg)
4 cos^2 theta + 4 cos theta = 1
i believe the quadratic formula applies but am stuck.
Log On
Question 955454: solve for solutions over the interval (0 deg, 360 deg)
4 cos^2 theta + 4 cos theta = 1
i believe the quadratic formula applies but am stuck. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! solve for solutions over the interval (0 deg, 360 deg)
4 cos^2 theta + 4 cos theta - 1 = 0
----
cos(t) = [-4 +- sqrt(16-4*4*-1)]/8 = [-4+-4sqrt(2)]/8
----
cos(t) = [-4 + 4sqrt(2)]/8 = [-1+sqrt(2)]/2 = 0.2071
or
cos(t) [-1-sqrt(2)]/2 = -1.2071 (extraneous)
-----------------------
t = arccos(0.2071) = 78.04 degrees or 281.96 degrees
---------
Cheers,
Stan H.
--------------
(