How do i simplify this?

Question

need to find the solution on the interval [0, 2pi), but i dont know how to use the trig identities to simplify the equations so that i can find an answer, help plz!

1) csc(x) + cot(x) = 1

2) [1 + cos(x)] / [1 - cos(x)] = 0

Answer 1

For all real numbers x and k is an element of the set of integers.

(1)

csc x +cot x =1

1/(sin x ) + (cos x)/(sin x) = 1

(1 + cos x)/sin x =1

(1 + cos x)²/sin² x = 1

(1 + cos x)²/(1 - cos² x) = 1, since sin² x = 1 - cos² x

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

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

1 + cos x = 1 - cos x

cos x = -cos x

2cos x = 0

cos x = 0

x = π/2 + kπ

0 ≤ (2k + 1)π/2 < 2π

{ π/2, 3π/2 }


(2)

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

Since 1 - cos x ≠ 0, cos x ≠ 1

Therefore, 1 + cos x = 0

cos x = -1

x = π + 2kπ

0 ≤ (2k + 1)π < 2π

{ π }

Answer 2

1)

csc(x) + cot(x) = 1

1/sin(x) + cos(x)/sin(x)=1

sin x != 0 (!= means not equal to)

So x != k*pi with k an integer

Muliply both sides by sin(x)

1+cos(x)=sin(x)

square both sides. We know automatically that sin(x)> 0 since 0<=1+cos(x)=sin(x) ; thus 2*k*pi < x < 3*k*pi ; k an integer

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

So either cos(x) = 0 or (1+cos(x)) = 0

If cos(x) = 0 then x = pi / 2 + k* pi ; k an integer

If (1+cos(x))=0 then cos(x)=-1 but that means sin(x) = 0 which isnt a valid answer since we initially said sin(x)!=0 (or we would be dividing by 0)

So all our solutions are

x = pi/2 + k*pi ; k an integer with our first constraint that sin(x) > 0 though we have
x = pi/2 + 2*k*pi

on [0,2*pi) our solutions are x = pi/2

2)

Clearly 1-cos(x) !=0 so that cos(x)!=1 so that x != 2*k*pi ; for k an integer

Clearly then this can only be 0 if the numerator is 0.

Thus 1+cos(x)=0.

So cos(x)=-1

So x = 2*k*pi + pi

So that on [0,2*pi) our solution is

x = pi

Answer 3

1.Let's write the first one
csc(x) +cot(x) = 1
so
1/sinx + cosx/sinx =1
In multiplying the 2 terms by sinx :
1+cosx= sinx
So (1+cosx)²= sinx²=1-cos²x
So 1+cos²x+2cosx =1 - cos²x
So 2 cos²x+2 cosx = 0
cosx( cosx +1) = 0
cosx= 0 x= pi/2 or x=-pi/2
cosx=-1 x= pi but impossible sinx =0 and equation not defined
Let's try the solutions in the first equation only one is correct because we took the square.
So 1 solution x= pi/2

2/ If cox is not equal to 1 we can multiply the 2 sides by cosx-1 so
1+cosx =0
cos x = -1
x= pi

Answer 4

1)cosec x + cot x =1

=> 1/sin x + cos x / sin x =1
=> (1 + cos x ) / sin x = 1
=> 1 + cos x = sin x

Squaring both sides :

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

Therefore , x = ╥ (pi)

Answer 5

1.
1/sin(x) + cos(x)/sin(x) = 1
(1 + cos(x)) / sin(x) = 1
1 + cos(x) = sin(x)
(1 + cos(x))^2 = (sin(x)) ^2
(1 + cos(x))^2 = 1 - (cos(x))^2
(cos(x))^2 + 2cos(x) + 1 = 1 - (cos(x))^2
(cos(x))^2 + 2cos(x) = - (cos(x))^2
2(cos(x))^2 + 2cos(x) = 0
cos(x) + 1 = 0
cos(x) = -1
... this didn't work. I did something wrong. Ignore this.


2.
1 + cos(x) = 0 (1 - cos(x)) where 1-cos(x) <> 0
1 + cos(x) = 0
cos(x) = -1
x = -5(pi), -3(pi), pi, 3(pi), 5(pi), ...

Answer 6

Try the answers in the back of the book!

Answer 7

sorry but i really dont feel like thinking right now

Answer 8

leave it for high schl kids