Question

4cos x-4 sin ^2 x+5=0

Answer 1

\[0 \le x \le \pi\]

Answer 2

convert sin^2 x to 1-cos^2 x, to get the cos all terms

Answer 3

so then it'd be 4cos x-4-4cos^2x?

Answer 4

4cosx - 4 sin^2 x + 5 = 0
4cosx - 4(1-cos^2 x) + 5 = 0
4cosx - 4 + 4cos^2 x + 5 = 0
4cos^2 x + 4cosx + 1 = 0

Answer 5

factor out this,
(2cosx + 1)(2cosx + 1) = 0

Answer 6

find the zeroes by take the factors be zero
2cosx + 1 = 0

Answer 7

so i split the 2cosx and the 1 and solve for both?

Answer 8

2cosx + 1 = 0
2cosx = -1
cosx = -1/2
yes, solve for x's

Answer 9

The answer to the question is 2pi/3

Answer 10

yeah, in interval 0≤x≤π
the solution only satisfied for x=2pi/3

Answer 11

so how did we go from -1/2 to 2pi/3? I'm missing a mental step somewhere

Answer 12

u have to memorize the values fo sine, cosine, etc... for the special angles ::)

Answer 13

Well yuck. Thanks!