The answer has to be in interval notation.?

Question

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

x=________ pi


Note: the answer must be an integer or a fraction.

I want to know if your answer is the same as mine. Thanks in advance.

Answer 1

Geofft has taken you half the distance till formation of the quadratic equation. I shall try and take you further to the answer you seek!!

Reproducing the correct section from Geofft

transform sin^2 x to 1-cos^2 x
so we have
2 - 2 cos^2 x - 5 cos x - 4 = 0
-2 cos^2 x - 5 cos x - 2 = 0
2 cos^2 x + 5 cos x + 2 = 0

this is a quadratic in terms of cos x (replace, say, u=cos x, and we have 2u^2 + 5u + 2 = 0).

Now further,

Solving for the quadratic gets you, (2u+1)*(u+2) = 0
So u = -0.5 or -2. Now since u is cosx it cannot be -2.
so u i.e. cosx = -0.5

Hence x = pi-pi/3 or pi+pi/3 i.e. 2pi/3 or 4pi/3. Now if we add 2pi to each of the answer we will get the next set of answers for which cosx = -0.5. This can be repetitive at 2pi interval.

Hope you got the answer!!!

Answer 2

not that bad. transform sin^2 x to 1-cos^2 x

so we have
2 - 2 cos^2 x - 5 cos x - 4 = 0
-2 cos^2 x - 5 cos x - 2 = 0
2 cos^2 x + 5 cos x + 2 = 0

this is a quadratic in terms of cos x (replace, say, u=cos x, and we have 2u^2 + 5u + 2 = 0). then quadratic-formula that and take the arccosine.

if that's how you got it, your answer's right. (or you could just plug in your answer...)

not sure what the "interval notation" means. there's no interval here.

Answer 3

2(sin x)^2 - 5 cos x - 4 = 0
2(1-(cos x)^2) - 5 cos x - 4 = 0
2+2(cos x)^2 + 5 cos x + 4 = 0
2(cos x)^2 + 5 cos x + 2 = 0
cos(x) = (-5 ± √(25 -16))/4
cos(x) = (-5 ± 3)/4
cos(x) = (-2,-1/2)
cos(x) = -1/2
x = ±2π/3 ± 2nπ
x = (± 2(1 ± 3n)/3)π

Answer 4

the equation you provided on solving gives
2(cosx)^2+5cosx+2=0
from which we get
cos x=-1/2 or cos x=-2
as cos x#-2, the only one solution is
cos x=-1/2
so x=2/3 pi i.e., in between (0,2pi)
i dont know about the interval notation. i am sorry