Question

If 0 ≤ x ≤ 2pi and 4 sin^2x + 4 cos x -1 = 0, which of the following sets contains all values of x?
(a) (pi/3, pi/6)
(b) (pi/3, 2pi/3)
(c) (2pi/3, -2pi/3)
(d) (pi/3, -pi/3)
(e) (2pi/3, 4pi/3)

Answer 1

rewrite
\[4\sin^2(x)+4\cos(x)-1=0\] as
\[4(1-\cos^2(x))+4\cos(x)-1=0\] and solve the quadratic equation in \(\cos(x)\]

Answer 2

you get
\[4-4\cos^2(x)+4\cos(x)-1=0\]
\[-4\cos^2(x)+4\cos(x)+3=0\] or
\[4\cos^2(x)-4\cos(x)+3=0\]

Answer 3

oops i mean
\[4\cos^2(x)-4\cos(x)-3=0\]

Answer 4

by some miracle this factors as
\[(2\cos(x)+1)(2\cos(x)-3)=0\] you can ignore the second factor because it is impossible, solve
\[2\cos(x)+1=0\]
\[\cos(x)=-\frac{1}{2}\] and finally solve for \(x\)

Answer 5

120

Answer 6

@satellite73 which answer choice is it then? apparently there's two solutions

Answer 7

i got 2pi/3 as one but what's the other?

Answer 8

oh i didn't solve it hold on

Answer 9

i think it is the last one
\[\frac{2\pi}{3}\] and
\[\frac{4\pi}{3}\] both these will give \(-\frac{1}{2}\) for cosine

Answer 10

ok thanks