can anyone solve these tricky trigonometric inequalitys?

Question

x is located 0< x <2pie

1. cos^2x-cosx-2=0
2.4sin^2x=1
3. 2cos^2x+3cosx-2=0

Answer 1

Why are they tricky?

Seems pretty simple to me.

1 - 16
2 - KY
3 - Punt

Answer 2

These are equations, by the way; can't see how they're inequalities.

1) cos^2(x) - cos(x) - 2 = 0

To solve this question, note that this is a quadratic in disguise. To make it more obvious, let u = cos(x). Then our equation is

u^2 - u - 2 = 0

Factoring, we get

(u - 2)(u + 1) = 0

Therefore, u = 2, u = -1, so our two equations are

cos(x) = 2, cos(x) = -1

cos(x) = 2 will have no solution, because the range of cos(x) is between -1 and 1. We reject this equation.

cos(x) = -1, however, implies that
x = pi.

2. 4sin^2(x) = 1.

Divide both sides by 4,

sin^2(x) = 1/4

Take the square root of both sides,

sin(x) = +/- (1/2)

Our two equations are sin(x) = 1/2, sin(x) = -1/2.

If sin(x) = 1/2, this occurs at the points pi/6 and 5pi/6.
If sin(x) = -1/2, this occurs at the points 7pi/6 and 11pi/6.
Therefore,
x = {pi/6, 5pi/6, 7pi/6, 11pi/6}

3. 2cos^2(x) + 3cos(x) - 2 = 0

Again, we factor this like a quadratic.

(2cos(x) + 1) (cos(x) - 2) = 0

2cos(x) + 1 = 0
cos(x) - 2 = 0 {which we can reject, for the same reason}.

2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = -1/2, which occurs at the points
x = {2pi/3, 4pi/3}

Answer 3

1) call cos x= z So z^2-z-2=0 and z = ((1-+3))/2

z= 2 and z= -1 cos x= 2 can not be So cosx =-1 and x=pi
2) sinx =+-1/2

sinx =1/2 x=pi/6 and x=5pi/6

sin x=-1/2 x=7pi/6 x= 11pi/6.

3) cos x= -2 ( can´t be) and cosx = 1/2

x=pi/3 and x= 5pi/3