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Popular Trigonometry >

cos(x)=(cot^2(x))/(csc^2(x))

  • Pre Algebra
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Solution

cos(x)=csc2(x)cot2(x)​

Solution

x=2π​+2πn,x=23π​+2πn
+1
Degrees
x=90∘+360∘n,x=270∘+360∘n
Solution steps
cos(x)=csc2(x)cot2(x)​
Subtract csc2(x)cot2(x)​ from both sidescos(x)−csc2(x)cot2(x)​=0
Simplify cos(x)−csc2(x)cot2(x)​:csc2(x)csc2(x)cos(x)−cot2(x)​
cos(x)−csc2(x)cot2(x)​
Convert element to fraction: cos(x)=csc2(x)cos(x)csc2(x)​=csc2(x)cos(x)csc2(x)​−csc2(x)cot2(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=csc2(x)cos(x)csc2(x)−cot2(x)​
csc2(x)csc2(x)cos(x)−cot2(x)​=0
g(x)f(x)​=0⇒f(x)=0csc2(x)cos(x)−cot2(x)=0
Express with sin, cos
−cot2(x)+cos(x)csc2(x)
Use the basic trigonometric identity: cot(x)=sin(x)cos(x)​=−(sin(x)cos(x)​)2+cos(x)csc2(x)
Use the basic trigonometric identity: csc(x)=sin(x)1​=−(sin(x)cos(x)​)2+cos(x)(sin(x)1​)2
Simplify −(sin(x)cos(x)​)2+cos(x)(sin(x)1​)2:sin2(x)−cos2(x)+cos(x)​
−(sin(x)cos(x)​)2+cos(x)(sin(x)1​)2
(sin(x)cos(x)​)2=sin2(x)cos2(x)​
(sin(x)cos(x)​)2
Apply exponent rule: (ba​)c=bcac​=sin2(x)cos2(x)​
cos(x)(sin(x)1​)2=sin2(x)cos(x)​
cos(x)(sin(x)1​)2
(sin(x)1​)2=sin2(x)1​
(sin(x)1​)2
Apply exponent rule: (ba​)c=bcac​=sin2(x)12​
Apply rule 1a=112=1=sin2(x)1​
=sin2(x)1​cos(x)
Multiply fractions: a⋅cb​=ca⋅b​=sin2(x)1⋅cos(x)​
Multiply: 1⋅cos(x)=cos(x)=sin2(x)cos(x)​
=−sin2(x)cos2(x)​+sin2(x)cos(x)​
Apply rule ca​±cb​=ca±b​=sin2(x)−cos2(x)+cos(x)​
=sin2(x)−cos2(x)+cos(x)​
sin2(x)cos(x)−cos2(x)​=0
g(x)f(x)​=0⇒f(x)=0cos(x)−cos2(x)=0
Solve by substitution
cos(x)−cos2(x)=0
Let: cos(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=cos(x)cos(x)=0,cos(x)=1
cos(x)=0,cos(x)=1
cos(x)=0:x=2π​+2πn,x=23π​+2πn
cos(x)=0
General solutions for cos(x)=0
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=2π​+2πn,x=23π​+2πn
x=2π​+2πn,x=23π​+2πn
cos(x)=1:x=2πn
cos(x)=1
General solutions for cos(x)=1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=0+2πn
x=0+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn,x=2πn
Since the equation is undefined for:2πnx=2π​+2πn,x=23π​+2πn

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Popular Examples

2tan^2(x)-3cot^2(x)=5solvefor x,13y=cos^4(1-2x)cos(x)+cos^2(x)+cos^3(x)=0cos(x)-sin(x)= 1/((sin(x)))-1/((cos(x)))sin^2(x)+cos^5(x)=2

Frequently Asked Questions (FAQ)

  • What is the general solution for cos(x)=(cot^2(x))/(csc^2(x)) ?

    The general solution for cos(x)=(cot^2(x))/(csc^2(x)) is x= pi/2+2pin,x=(3pi)/2+2pin
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