sin^2(x)+cos(x)-1=0, x=?

Question

Can someone please help me with this problem... I have no clue, so if you can show the steps, that would be great.

Answer 1

For this problem, you use the trigonometric identity:

sin^2(x) + cos(x) - 1 = 0
we know that sin^2(x) + cos^2(x) = 1 [pythagorean identity]
so, sin^2(x) = 1 - cos^2(x)

now substituting this expression for sin^2(x) into the original function

1 - cos^2(x) + cos(x) -1 = 0
-cos^2(x) + cos(x) = 0
factorizing gives:

cos(x)[1-cos(x)] = 0
therefore: cos(x) = 0 or 1 - cos(x) = 0

for cos(x) = 0
x = arcos 0
x = 90
x = pi/2, -pi/2 (which is 3pi/2)

for 1 - cos(x) = 0
1 = cos(x)
x = arcos 1
x = 0

So you have three solutions that are:
x = 0, pi/2, 3pi/2 (where 0 < x < 2pi)

Hope this helps! =)

Answer 2

Start by using the trigonometric identity sin^2(x) + cos^2(x) = 1.
Solving this for sin^2(x) gives sin^2(x) = 1 - cos^2(x). Now, you can substitute this into the original equation.

Substituting sin^2(x) = 1 - cos^2(x) into sin^2(x)+cos(x)-1=0 yields:

1 - cos^2(x) + cos(x) - 1 = 0. The 1's cancel out and you're left with:

cos(x) - cos^2(x) = 0. Factor this out to get

cos(x)(1 - cos(x)) = 0. So this equation is valid when cos(x) = 0 and when 1 - cos(x) = 0.

For cos(x) = 0, x = pi/2 and -pi/2
For 1 - cos(x) = 0 => cos(x) = 1, x = 0

So there are 3 solutions to the equation. Finally plug them all into the original equation to make sure they all work (they do).

EDIT: Upon submittal, I realized there was more than one solution, and redid my derivation. I'm not sure of your math ability but to avoid any confusion, -pi/2 is in the same spot on the unit circle as 3pi/2 and they are both valid solutions.

Answer 3

Flashy and Thugster are both almost correct.

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

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


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

cos(x)*( 1 - cos(x)) = 0,

so cos(x) = 1 or 0. The only solutions on the circle 0<= x < 2 pi

are

x = 0, pi/2, and 3*pi/2. In degrees, these are 0, 90, and 270.

Answer 4

Let's assume you want solutions in the interval 0 <= x < 2pi.

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

First, use the fact that sin^2(x) = 1 - cos^2(x)

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

Note that the 1 and -1 cancel out, giving us

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

Factor out cos(x),

cos(x) [-cos(x) + 1] = 0

This gives us two equations:

cos(x) = 0, and
-cos(x) + 1 = 0

Therefore,

cos(x) = 0
cos(x) = 1

All we have to do is individually solve these.

cos(x) is equal to 0 at the points x = {pi/2, 3pi/2}
cos(x) is equal to 1 at the points x = {0}

Therefore, our solutions are

x = {pi/2, 3pi/2, 0}

Answer 5

sin2x - cosx - 2sinx +1 = 0 thus, 2sinxcosx - cosx - 2sinx +1 = 0 ( as sin2x = 2sinxcosx ) 2sinx ( cosx - 1 ) - 1 ( cosx - 1 ) = 0 ( by taking out 2sinx and -1 common...) ( cosx - 1 ) ( 2sinx - 1 ) = 0 ( by taking out ( cosx - 1 ) common....) thus, either ( cosx - 1 ) = 0 or ( 2sinx - 1 ) = 0 cosx = 1 or sinx = 1/2 thus, x = 0 or x = 30 degree

Answer 6

i get it!!!

sin x2 equals the cost of losing your soul (known as -1) which equals (=) 0

so X (me) = 0

Answer 7

HAHAHAHAHA!!!it's so funny whenever i see math..

Answer 8

maybe you should ask your teacher.............