Geometry question.?

Question

Find all values of x such that cos^2x - cosx = sin^2x, where 0 less than equal to x less than equal to 2-pi. (0 > x > 2pi)

Can someone refresh me how to attempt these kinds of geometry questions? It's part of my readiness test for university math.

Answer 1

cos^2x - sin^2x = cosx

cos2x = cosx

This is possible for x = 0, 120 degrees (i.e. 2pi/3)

Alternatively, u can do it as

sin^2x - 1 - cos^2x

so, your equation is
cos^2x - cosx = sin^2x

2cos^2x - cosx - 1=0

put cosx = y

so, 2y^2 - y - 1 = 0

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

y = 1 and -1/2

cosx = 1 and -1/2

so, x = 0 and 120 degrees or 0 and 2pi/3

since your interval for x does not include 0, the only soln is 2pi/3

Answer 2

1

Answer 3

if you solve this problem you find two answers

Cos x = 1 and Cos x = - 0.5

Since Cos x = 1 will mean x = 0 deg

the answer will be x = Cos Inverse of - 0.5 which is 120 deg or 2Pi / 3
p.s. you can solve the problem by replacing sin^x by (1-cos^2x)

Then use simple algebra and factorise

Answer 4

basically, the only thing that i can find in common with sacred geometry and alchemy is that they both deal with types of spirituality. Sacred geometry deals with the use of geometric patterns and shapes in the design and architecture of churches, temples, mosques or any place of spiritual gathering. These specific shapes mean certain things, depending on the religion or belief system. For example, the pentagram in Christianity refers to the five senses. In Catholicism, it can sometimes refer to the Devil, or the Evil One. Alchemy deals with the process of turning metals into gold, using chemicals and science. However, Alchemy is also a belief in finding the ultimate wisdom through chemistry. Therefore, alchemy is a spirituality all its own. All in all, the only thing that I can see in common between the two is that they both deal with a type, or types, of spirituality.

Answer 5

change sin^2x into 1 - cos^2x (pythagorean) and move everything over to one side. you now have a quadratic equation with just cos. Also, I think your interval is backwards. It should be 0

Answer 6

cos²(x) - cos(x) = sin²(x) [Substitute sin²(x) = 1 - cos²(x)]
cos²(x) - cos(x) = 1 - cos²(x) [Add cos²(x) - 1 to both sides]
2cos²(x) - cos(x) - 1 = 0 [Factor]
(2cos(x) + 1) · (cos(x) - 1) = 0 [Set each factor equal to zero]
2cos(x) + 1 = 0 or cos(x) - 1 = 0 [Solve each]

If 2cos(x) + 1 = 0, then
2cos(x) = -1
cos(x) = -1 / 2
x = arccos(-1 / 2)
x = 2π / 3
Over the interval 0 ≤ x ≤ 2π,
x = 2π / 3 or 4π / 3.

If cos(x) - 1 = 0, then
cos(x) = 1
x = arccos(1)
x = 0
Over the interval 0 ≤ x ≤ 2π,
x = 0 or 2π.

All solutions, then, are 0, 2π / 3, 4π / 3, or 2π.

Answer 7

if by this you mean

cos(x)^2 - cos(x) = sin(x)^2

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

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

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

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

2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = (-1/2)
x = 120° or 240°

cos(x) - 1 = 0
cos(x) = 1
x = 0° or 360°

x = 0°, 120°, 240°, or 360°

ANS : 0°, ((2pi)/3), ((4pi)/3), or 2pi

Also you have your inequality wrote wrong, it should read

0 <= x <= 2pi

Answer 8

(cosx)^2 - cosx = (sinx)^2. Put c = cosx. Then c^2 -c = 1 -c^2, ie.,
2c^2 -c -1 +0, ie., (2c+1)(c-1)=0 & c = -(1/2)..(i) or c = 1..(ii). Now, for (i)
holding, x =pi (+/-)pi/3, ie., x =2pi/3 or 4pi/3. For (ii) holding, x = 0 or 2pi.
Question poser states (a) "where 0 less than equal to x less than equal to 2-pi" & then (b) "(0 > x > 2pi)", contradicting (a). Obviously (b) should
read "(0 sol'n set is x = 0, 2pi/3, 4pi/3, 2pi for x in [0,2pi].

Answer 9

12 i think

Answer 10

what property justifies the statement 5(7z-3y)=35z-15y