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Popular Trigonometry >

sqrt(1+cot^2(x))=8

  • Pre Algebra
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Solution

1+cot2(x)​=8

Solution

x=0.12532…+πn,x=3.01626…+πn
+1
Degrees
x=7.18075…∘+180∘n,x=172.81924…∘+180∘n
Solution steps
1+cot2(x)​=8
Solve by substitution
1+cot2(x)​=8
Let: cot(x)=u1+u2​=8
1+u2​=8:u=37​,u=−37​
1+u2​=8
Square both sides:1+u2=64
1+u2​=8
(1+u2​)2=82
Expand (1+u2​)2:1+u2
(1+u2​)2
Apply radical rule: a​=a21​=((1+u2)21​)2
Apply exponent rule: (ab)c=abc=(1+u2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1+u2
Expand 82:64
82
82=64=64
1+u2=64
1+u2=64
Solve 1+u2=64:u=37​,u=−37​
1+u2=64
Move 1to the right side
1+u2=64
Subtract 1 from both sides1+u2−1=64−1
Simplifyu2=63
u2=63
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=63​,u=−63​
63​=37​
63​
Prime factorization of 63:32⋅7
63
63divides by 363=21⋅3=3⋅21
21divides by 321=7⋅3=3⋅3⋅7
3,7 are all prime numbers, therefore no further factorization is possible=3⋅3⋅7
=32⋅7
=32⋅7​
Apply radical rule: =7​32​
Apply radical rule: 32​=3=37​
−63​=−37​
−63​
63​=37​
63​
Prime factorization of 63:32⋅7
63
63divides by 363=21⋅3=3⋅21
21divides by 321=7⋅3=3⋅3⋅7
3,7 are all prime numbers, therefore no further factorization is possible=3⋅3⋅7
=32⋅7
=32⋅7​
Apply radical rule: =7​32​
Apply radical rule: 32​=3=37​
=−37​
u=37​,u=−37​
u=37​,u=−37​
Verify Solutions:u=37​True,u=−37​True
Check the solutions by plugging them into 1+u2​=8
Remove the ones that don't agree with the equation.
Plug in u=37​:True
1+(37​)2​=8
1+(37​)2​=8
1+(37​)2​
(37​)2=32⋅7
(37​)2
Apply exponent rule: (a⋅b)n=anbn=32(7​)2
(7​)2:7
Apply radical rule: a​=a21​=(721​)2
Apply exponent rule: (ab)c=abc=721​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=7
=32⋅7
=1+32⋅7​
32⋅7=63
32⋅7
32=9=9⋅7
Multiply the numbers: 9⋅7=63=63
=1+63​
Add the numbers: 1+63=64=64​
Factor the number: 64=82=82​
Apply radical rule: 82​=8=8
8=8
True
Plug in u=−37​:True
1+(−37​)2​=8
1+(−37​)2​=8
1+(−37​)2​
(−37​)2=32⋅7
(−37​)2
Apply exponent rule: (−a)n=an,if n is even(−37​)2=(37​)2=(37​)2
Apply exponent rule: (a⋅b)n=anbn=32(7​)2
(7​)2:7
Apply radical rule: a​=a21​=(721​)2
Apply exponent rule: (ab)c=abc=721​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=7
=32⋅7
=1+32⋅7​
32⋅7=63
32⋅7
32=9=9⋅7
Multiply the numbers: 9⋅7=63=63
=1+63​
Add the numbers: 1+63=64=64​
Factor the number: 64=82=82​
Apply radical rule: 82​=8=8
8=8
True
The solutions areu=37​,u=−37​
Substitute back u=cot(x)cot(x)=37​,cot(x)=−37​
cot(x)=37​,cot(x)=−37​
cot(x)=37​:x=arccot(37​)+πn
cot(x)=37​
Apply trig inverse properties
cot(x)=37​
General solutions for cot(x)=37​cot(x)=a⇒x=arccot(a)+πnx=arccot(37​)+πn
x=arccot(37​)+πn
cot(x)=−37​:x=arccot(−37​)+πn
cot(x)=−37​
Apply trig inverse properties
cot(x)=−37​
General solutions for cot(x)=−37​cot(x)=−a⇒x=arccot(−a)+πnx=arccot(−37​)+πn
x=arccot(−37​)+πn
Combine all the solutionsx=arccot(37​)+πn,x=arccot(−37​)+πn
Show solutions in decimal formx=0.12532…+πn,x=3.01626…+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for sqrt(1+cot^2(x))=8 ?

    The general solution for sqrt(1+cot^2(x))=8 is x=0.12532…+pin,x=3.01626…+pin
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