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Popular Trigonometry >

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

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Solution

1−cos(x)1​+1+cos(x)1​=2csc(x)

Solution

x=2π​+2πn
+1
Degrees
x=90∘+360∘n
Solution steps
1−cos(x)1​+1+cos(x)1​=2csc(x)
Subtract 2csc(x) from both sides(−cos(x)+1)(cos(x)+1)2cos2(x)csc(x)−2csc(x)+2​=0
g(x)f(x)​=0⇒f(x)=02cos2(x)csc(x)−2csc(x)+2=0
Express with sin, cos2cos2(x)sin(x)1​−2⋅sin(x)1​+2=0
Simplify 2cos2(x)sin(x)1​−2⋅sin(x)1​+2:sin(x)2cos2(x)−2+2sin(x)​
2cos2(x)sin(x)1​−2⋅sin(x)1​+2
2cos2(x)sin(x)1​=sin(x)2cos2(x)​
2cos2(x)sin(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=sin(x)1⋅2cos2(x)​
Multiply the numbers: 1⋅2=2=sin(x)2cos2(x)​
2⋅sin(x)1​=sin(x)2​
2⋅sin(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=sin(x)1⋅2​
Multiply the numbers: 1⋅2=2=sin(x)2​
=sin(x)2cos2(x)​−sin(x)2​+2
Combine the fractions sin(x)2cos2(x)​−sin(x)2​:sin(x)2cos2(x)−2​
Apply rule ca​±cb​=ca±b​=sin(x)2cos2(x)−2​
=sin(x)2cos2(x)−2​+2
Convert element to fraction: 2=sin(x)2sin(x)​=sin(x)2cos2(x)−2​+sin(x)2sin(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin(x)2cos2(x)−2+2sin(x)​
sin(x)2cos2(x)−2+2sin(x)​=0
g(x)f(x)​=0⇒f(x)=02cos2(x)−2+2sin(x)=0
Subtract 2sin(x) from both sides2cos2(x)−2=−2sin(x)
Square both sides(2cos2(x)−2)2=(−2sin(x))2
Subtract (−2sin(x))2 from both sides(2cos2(x)−2)2−4sin2(x)=0
Factor (2cos2(x)−2)2−4sin2(x):4(cos2(x)−1+sin(x))(cos2(x)−1−sin(x))
(2cos2(x)−2)2−4sin2(x)
Rewrite (2cos2(x)−2)2−4sin2(x) as (2cos2(x)−2)2−(2sin(x))2
(2cos2(x)−2)2−4sin2(x)
Rewrite 4 as 22=(2cos2(x)−2)2−22sin2(x)
Apply exponent rule: ambm=(ab)m22sin2(x)=(2sin(x))2=(2cos2(x)−2)2−(2sin(x))2
=(2cos2(x)−2)2−(2sin(x))2
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(2cos2(x)−2)2−(2sin(x))2=((2cos2(x)−2)+2sin(x))((2cos2(x)−2)−2sin(x))=((2cos2(x)−2)+2sin(x))((2cos2(x)−2)−2sin(x))
Refine=(2cos2(x)+2sin(x)−2)(2cos2(x)−2sin(x)−2)
Factor 2cos2(x)−2+2sin(x):2(cos2(x)−1+sin(x))
2cos2(x)−2+2sin(x)
Factor out common term 2=2(cos2(x)−1+sin(x))
=2(cos2(x)+sin(x)−1)(2cos2(x)−2sin(x)−2)
Factor 2cos2(x)−2−2sin(x):2(cos2(x)−1−sin(x))
2cos2(x)−2−2sin(x)
Factor out common term 2=2(cos2(x)−1−sin(x))
=2(cos2(x)−1+sin(x))⋅2(cos2(x)−1−sin(x))
Refine=4(cos2(x)−1+sin(x))(cos2(x)−1−sin(x))
4(cos2(x)−1+sin(x))(cos2(x)−1−sin(x))=0
Solving each part separatelycos2(x)−1+sin(x)=0orcos2(x)−1−sin(x)=0
cos2(x)−1+sin(x)=0:x=2πn,x=π+2πn,x=2π​+2πn
cos2(x)−1+sin(x)=0
Rewrite using trig identities
−1+cos2(x)+sin(x)
Use the Pythagorean identity: 1=cos2(x)+sin2(x)1−cos2(x)=sin2(x)=sin(x)−sin2(x)
sin(x)−sin2(x)=0
Solve by substitution
sin(x)−sin2(x)=0
Let: sin(x)=uu−u2=0
u−u2=0:u=0,u=1
u−u2=0
Write in the standard form ax2+bx+c=0−u2+u=0
Solve with the quadratic formula
−u2+u=0
Quadratic Equation Formula:
For a=−1,b=1,c=0u1,2​=2(−1)−1±12−4(−1)⋅0​​
u1,2​=2(−1)−1±12−4(−1)⋅0​​
12−4(−1)⋅0​=1
12−4(−1)⋅0​
Apply rule 1a=112=1=1−4(−1)⋅0​
Apply rule −(−a)=a=1+4⋅1⋅0​
Apply rule 0⋅a=0=1+0​
Add the numbers: 1+0=1=1​
Apply rule 1​=1=1
u1,2​=2(−1)−1±1​
Separate the solutionsu1​=2(−1)−1+1​,u2​=2(−1)−1−1​
u=2(−1)−1+1​:0
2(−1)−1+1​
Remove parentheses: (−a)=−a=−2⋅1−1+1​
Add/Subtract the numbers: −1+1=0=−2⋅10​
Multiply the numbers: 2⋅1=2=−20​
Apply the fraction rule: −ba​=−ba​=−20​
Apply rule a0​=0,a=0=−0
=0
u=2(−1)−1−1​:1
2(−1)−1−1​
Remove parentheses: (−a)=−a=−2⋅1−1−1​
Subtract the numbers: −1−1=−2=−2⋅1−2​
Multiply the numbers: 2⋅1=2=−2−2​
Apply the fraction rule: −b−a​=ba​=22​
Apply rule aa​=1=1
The solutions to the quadratic equation are:u=0,u=1
Substitute back u=sin(x)sin(x)=0,sin(x)=1
sin(x)=0,sin(x)=1
sin(x)=0:x=2πn,x=π+2πn
sin(x)=0
General solutions for sin(x)=0
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+2πn
sin(x)=1:x=2π​+2πn
sin(x)=1
General solutions for sin(x)=1
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=2π​+2πn
x=2π​+2πn
Combine all the solutionsx=2πn,x=π+2πn,x=2π​+2πn
cos2(x)−1−sin(x)=0:x=23π​+2πn,x=2πn,x=π+2πn
cos2(x)−1−sin(x)=0
Rewrite using trig identities
−1+cos2(x)−sin(x)
Use the Pythagorean identity: 1=cos2(x)+sin2(x)1−cos2(x)=sin2(x)=−sin(x)−sin2(x)
−sin(x)−sin2(x)=0
Solve by substitution
−sin(x)−sin2(x)=0
Let: sin(x)=u−u−u2=0
−u−u2=0:u=−1,u=0
−u−u2=0
Write in the standard form ax2+bx+c=0−u2−u=0
Solve with the quadratic formula
−u2−u=0
Quadratic Equation Formula:
For a=−1,b=−1,c=0u1,2​=2(−1)−(−1)±(−1)2−4(−1)⋅0​​
u1,2​=2(−1)−(−1)±(−1)2−4(−1)⋅0​​
(−1)2−4(−1)⋅0​=1
(−1)2−4(−1)⋅0​
Apply rule −(−a)=a=(−1)2+4⋅1⋅0​
(−1)2=1
(−1)2
Apply exponent rule: (−a)n=an,if n is even(−1)2=12=12
Apply rule 1a=1=1
4⋅1⋅0=0
4⋅1⋅0
Apply rule 0⋅a=0=0
=1+0​
Add the numbers: 1+0=1=1​
Apply rule 1​=1=1
u1,2​=2(−1)−(−1)±1​
Separate the solutionsu1​=2(−1)−(−1)+1​,u2​=2(−1)−(−1)−1​
u=2(−1)−(−1)+1​:−1
2(−1)−(−1)+1​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅11+1​
Add the numbers: 1+1=2=−2⋅12​
Multiply the numbers: 2⋅1=2=−22​
Apply the fraction rule: −ba​=−ba​=−22​
Apply rule aa​=1=−1
u=2(−1)−(−1)−1​:0
2(−1)−(−1)−1​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅11−1​
Subtract the numbers: 1−1=0=−2⋅10​
Multiply the numbers: 2⋅1=2=−20​
Apply the fraction rule: −ba​=−ba​=−20​
Apply rule a0​=0,a=0=−0
=0
The solutions to the quadratic equation are:u=−1,u=0
Substitute back u=sin(x)sin(x)=−1,sin(x)=0
sin(x)=−1,sin(x)=0
sin(x)=−1:x=23π​+2πn
sin(x)=−1
General solutions for sin(x)=−1
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=23π​+2πn
x=23π​+2πn
sin(x)=0:x=2πn,x=π+2πn
sin(x)=0
General solutions for sin(x)=0
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+2πn
Combine all the solutionsx=23π​+2πn,x=2πn,x=π+2πn
Combine all the solutionsx=2πn,x=π+2πn,x=2π​+2πn,x=23π​+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into 1−cos(x)1​+1+cos(x)1​=2csc(x)
Remove the ones that don't agree with the equation.
Check the solution 2πn:False
2πn
Plug in n=12π1
For 1−cos(x)1​+1+cos(x)1​=2csc(x)plug inx=2π11−cos(2π1)1​+1+cos(2π1)1​=2csc(2π1)
Undefined
⇒False
Check the solution π+2πn:False
π+2πn
Plug in n=1π+2π1
For 1−cos(x)1​+1+cos(x)1​=2csc(x)plug inx=π+2π11−cos(π+2π1)1​+1+cos(π+2π1)1​=2csc(π+2π1)
Undefined
⇒False
Check the solution 2π​+2πn:True
2π​+2πn
Plug in n=12π​+2π1
For 1−cos(x)1​+1+cos(x)1​=2csc(x)plug inx=2π​+2π11−cos(2π​+2π1)1​+1+cos(2π​+2π1)1​=2csc(2π​+2π1)
Refine2=2
⇒True
Check the solution 23π​+2πn:False
23π​+2πn
Plug in n=123π​+2π1
For 1−cos(x)1​+1+cos(x)1​=2csc(x)plug inx=23π​+2π11−cos(23π​+2π1)1​+1+cos(23π​+2π1)1​=2csc(23π​+2π1)
Refine2=−2
⇒False
x=2π​+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 1/(1-cos(x))+1/(1+cos(x))=2csc(x) ?

    The general solution for 1/(1-cos(x))+1/(1+cos(x))=2csc(x) is x= pi/2+2pin
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