sin^2 = 1-cos^2, so
4 + 4cos = 1 - cos^2
cos^2 + 4cos + 3 = 0
(cosx + 1)(cosx + 3) = 0
cosx = -1 or -3
-3 is out, so cosx = -1.
x = 180 or -180
2sinx = 5cosx
tanx = 2/5
x ≈ 21.8°
tanx is positive in QI and QIII, so our solutions are
x = 21.8° or -158.2°
Solve the following equations for -180《 x《 180
A) 4+4cosx=sin^2x
B) 2sin=5cosx
Thankyou
1 answer
1 answer