can someone solve cosQ-sin^2Q+cos^2Q=0 for Q (Q is theta)?

Question

Be specific please i really want to know

Answer 1

cos(Q) - sin(Q)^2 + cos(Q)^2 = 0
cos(Q) - (1 - cos(Q)^2) + cos(Q)^2 = 0
cos(Q) - 1 + cos(Q)^2 + cos(Q)^2 = 0
2cos(Q)^2 + cos(Q) - 1 = 0
(2cos(Q) - 1)(cos(Q) + 1) = 0

2cos(Q) - 1 = 0
2cos(Q) = 1
cos(Q) = (1/2)
Q = (pi/3) or (5/3)pi

cos(Q) + 1 = 0
cos(Q) = -1
Q = pi

ANS : (pi/3), pi, or (5/3)pi

Answer 2

cos Θ-sin^2 Θ+cos^2 Θ=0 sin^2=-1cos^2 so
cos Θ-(1-cos^2 Θ)+cos^2 Θ
2cos^2 Θ+cos Θ-1=0
(2 cos Θ -1)(cos Θ +1)
cos Θ=1/2
Θ=arccos 1/2=60°
cos Θ=-1
Θ=arccos -1=180°

Answer 3

Is this your question
cos(Q) - sin^2(Q) + cos^2(Q) = 0
or is this it
cos(Q) - sin^2(Q) - cos^2(Q) = 0

For the first I got three possibilities:
pi
1/3*pi
-1/3*pi

for the second I got:
0

The answer exists, and it can in fact be solved.

Answer 4

Wow.
No I can't lol.
I need math help myself!
I suck at math.. That's a BIG problem!!!