the value of x which satisfies the eq:( cos^2x-3cosx+2)/sin^2x=1 , lies in the open interval?

Question

]pi/6,pi/2[ & ]pi/4 ,pi/2[
how???????

Answer 1

ah such a cuite name you have nishi .well dear look what am doing ;
as you know
sin^2x + cos^2 = +1 so
sin^2x = 1-cos^2x formula one;
(cos^2x-3cosx+2) / sin^2x=1
cos^2x-3cosx+2 = (sin^2x)*1
cos^2x-3cosx+2 = sin^2x
{ here put formual (a) instead of sin^2x }
cos^2x-3cosx+2 = 1- (cos^2x)
cos^2x-3cosx+ cos^2x = 1-2
2cos^2x - 3cosx =-1
factor ( cosx)
cosx(2cosx - 3) = -1
cosx= -1 as you know in 0, 2pi cosx = +1 & in pi cosx = -1
Or
2 cosx -3 = -1 ; 2cosx = -1+3 ; 2cosx = 2 ; cosx = +1
x = 0 , x = 180 or pi ; x = 360 or 2pi
Good Luck sweetheart.

Answer 2

First, so some substituting:

sin^2 x = 1 - cos^2 x

So, (cos^2x - 3cosx + 2) / sin^2x = 1
or (cos^2x - 3cosx + 2) / (1 - cos^2 x) = 1

(cos^2x - 3cosx + 2) = (1 - cos^2 x)
2cos^2 x - 3cos x + 1 = 0

Substitute again, u = cos x
Then 2x^2 - 3x + 1 = 0
(2x - 1 )(x - 1) = 0
x = (1/2) or x = 1

Therefore,
cos x = 1/2
Since cos x > 0, the x-coordinate is positive so the solution is in both the 1st and 4th quadrants:

x = pi/3 and x = 2pi - pi/3 = 5pi/3

cos x = 1
Since cos x > 0, the x-coordinate is positive and the adjacent = the hytpotenuse, the solution is
x = 0

The possible answers are x = 0, x = pi/3 and x = 5pi/3
But since f(0) = 0/0, the answer is
x = pi/3 and x = 5pi/3

Answer 3

Since your rational expression equals 1, the numerator equals the denominator.

So cos^x - 3cosx + 2 = sin^2x

replace sin^2x with 1-cos^2x, and move these terms to the left to get

2 cos^2x - 3cosx + 1=0

Factor: (2 cosx - 1)(cosx-1)=0

So cos x = 1/2 or cosx = 1

x = pi/3 or 5pi/3 or 0

Discard 0 because the denominator would be 0

Answer 4

For,
[cos^2 x - 3cos x + 2] / sin^2 x =1
=> cos^2 x - 3cos x + 2 = sin^2 x = 1 - cos^2 x, (for non zero x)
=> 2cos^2 x - 3cos x + 1 = 0
then, (2cos x -1).(cos x -1) = 0
then cos x = 1/2 or cos x = 1
x= 2n.pi + pi/3 or 2n.pi
where n is an integer.
Alternatively factor the numerator and denominator in linear factors of cos x and cancel out, assuming x does not equal n.pi

PS-
-i.ln [x + V(x^2 -1)] = arccos x
(derived from the exponential definitions of trig functions)

Answer 5

Your equation factored is
((cosx-2)(cosx-1))/sin^2x=1 , then covert sin^2x to it's identity and we have
((cosx-2)(cosx-1))/(1-cos^2x)=
1, then factor the denominator giving
((cosx-2)(cosx-1))/
(1+cosx)(1-cosx)=1 and cancel cosx-1
and 1-cos in the numerator and denominator respectively
giving
(cosx-2)/-(1+cosx)=1
cosx -2 =-cosx -1
2cosx=1
cosx=1/2
x=pi/3 or -pi/3

Checking use cos(pi/3)=1/2 and sin(pi/3)=sqrt3/2
back into your equation and we have
(1/2)^2-1 1/2 +2)(2)(2)/sqrt3sqrt3=
(3/4)4/3=1 OK

Answer 6

assuming you mean

((cos(x))^2 - 3(cos(x)) + 2)/((sin(x))^2) = 1
((cos(x) - 2)(cos(x) - 1))/(1 - (cos(x))^2) = 1
((cos(x) - 2)(cos(x) - 1))/(-(cos(x))^2 + 1) = 1
((cos(x) - 2)(cos(x) - 1))/(-((cos(x))^2) - 1) = 1
-((cos(x) - 2)(cos(x) - 1))/((cos(x))^2 - 1) = 1
((cos(x) - 2)(cos(x) - 1))/((cos(x)^2 - 1) = -1
((cos(x) - 2)(cos(x) - 1))/((cos(x) - 1)(cos(x) + 1)) = -1

the (cos(x) - 1)s cancel out leaving you with

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

multiply both sides by cos(x) + 1

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

2cos(x) = 1

cos(x) = (1/2)

x = 60° or 300°

in radian form, that would be

x = (pi/3) or ((5pi)/3)

but only (pi/3) lies in the open intervals

ANS : (pi/3)

pi/6 = 30°
pi/2 = 90°

since 60° is in between 30° and 60°, (pi/3) works out

pi/4 = 45°
pi/2 = 90°

same reason above

Answer 7

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