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Popular Trigonometry >

5tan^4(x)-10tan^2(x)+1=0

  • Pre Algebra
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Solution

5tan4(x)−10tan2(x)+1=0

Solution

x=0.94247…+πn,x=−0.94247…+πn,x=0.31415…+πn,x=−0.31415…+πn
+1
Degrees
x=54∘+180∘n,x=−54∘+180∘n,x=18∘+180∘n,x=−18∘+180∘n
Solution steps
5tan4(x)−10tan2(x)+1=0
Solve by substitution
5tan4(x)−10tan2(x)+1=0
Let: tan(x)=u5u4−10u2+1=0
5u4−10u2+1=0:u=55+25​​​,u=−55+25​​​,u=55−25​​​,u=−55−25​​​
5u4−10u2+1=0
Rewrite the equation with v=u2 and v2=u45v2−10v+1=0
Solve 5v2−10v+1=0:v=55+25​​,v=55−25​​
5v2−10v+1=0
Solve with the quadratic formula
5v2−10v+1=0
Quadratic Equation Formula:
For a=5,b=−10,c=1v1,2​=2⋅5−(−10)±(−10)2−4⋅5⋅1​​
v1,2​=2⋅5−(−10)±(−10)2−4⋅5⋅1​​
(−10)2−4⋅5⋅1​=45​
(−10)2−4⋅5⋅1​
Apply exponent rule: (−a)n=an,if n is even(−10)2=102=102−4⋅5⋅1​
Multiply the numbers: 4⋅5⋅1=20=102−20​
102=100=100−20​
Subtract the numbers: 100−20=80=80​
Prime factorization of 80:24⋅5
80
80divides by 280=40⋅2=2⋅40
40divides by 240=20⋅2=2⋅2⋅20
20divides by 220=10⋅2=2⋅2⋅2⋅10
10divides by 210=5⋅2=2⋅2⋅2⋅2⋅5
2,5 are all prime numbers, therefore no further factorization is possible=2⋅2⋅2⋅2⋅5
=24⋅5
=24⋅5​
Apply radical rule: nab​=na​nb​=5​24​
Apply radical rule: nam​=anm​24​=224​=22=225​
Refine=45​
v1,2​=2⋅5−(−10)±45​​
Separate the solutionsv1​=2⋅5−(−10)+45​​,v2​=2⋅5−(−10)−45​​
v=2⋅5−(−10)+45​​:55+25​​
2⋅5−(−10)+45​​
Apply rule −(−a)=a=2⋅510+45​​
Multiply the numbers: 2⋅5=10=1010+45​​
Factor 10+45​:2(5+25​)
10+45​
Rewrite as=2⋅5+2⋅25​
Factor out common term 2=2(5+25​)
=102(5+25​)​
Cancel the common factor: 2=55+25​​
v=2⋅5−(−10)−45​​:55−25​​
2⋅5−(−10)−45​​
Apply rule −(−a)=a=2⋅510−45​​
Multiply the numbers: 2⋅5=10=1010−45​​
Factor 10−45​:2(5−25​)
10−45​
Rewrite as=2⋅5−2⋅25​
Factor out common term 2=2(5−25​)
=102(5−25​)​
Cancel the common factor: 2=55−25​​
The solutions to the quadratic equation are:v=55+25​​,v=55−25​​
v=55+25​​,v=55−25​​
Substitute back v=u2,solve for u
Solve u2=55+25​​:u=55+25​​​,u=−55+25​​​
u2=55+25​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=55+25​​​,u=−55+25​​​
Solve u2=55−25​​:u=55−25​​​,u=−55−25​​​
u2=55−25​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=55−25​​​,u=−55−25​​​
The solutions are
u=55+25​​​,u=−55+25​​​,u=55−25​​​,u=−55−25​​​
Substitute back u=tan(x)tan(x)=55+25​​​,tan(x)=−55+25​​​,tan(x)=55−25​​​,tan(x)=−55−25​​​
tan(x)=55+25​​​,tan(x)=−55+25​​​,tan(x)=55−25​​​,tan(x)=−55−25​​​
tan(x)=55+25​​​:x=arctan​55+25​​​​+πn
tan(x)=55+25​​​
Apply trig inverse properties
tan(x)=55+25​​​
General solutions for tan(x)=55+25​​​tan(x)=a⇒x=arctan(a)+πnx=arctan​55+25​​​​+πn
x=arctan​55+25​​​​+πn
tan(x)=−55+25​​​:x=arctan​−55+25​​​​+πn
tan(x)=−55+25​​​
Apply trig inverse properties
tan(x)=−55+25​​​
General solutions for tan(x)=−55+25​​​tan(x)=−a⇒x=arctan(−a)+πnx=arctan​−55+25​​​​+πn
x=arctan​−55+25​​​​+πn
tan(x)=55−25​​​:x=arctan​55−25​​​​+πn
tan(x)=55−25​​​
Apply trig inverse properties
tan(x)=55−25​​​
General solutions for tan(x)=55−25​​​tan(x)=a⇒x=arctan(a)+πnx=arctan​55−25​​​​+πn
x=arctan​55−25​​​​+πn
tan(x)=−55−25​​​:x=arctan​−55−25​​​​+πn
tan(x)=−55−25​​​
Apply trig inverse properties
tan(x)=−55−25​​​
General solutions for tan(x)=−55−25​​​tan(x)=−a⇒x=arctan(−a)+πnx=arctan​−55−25​​​​+πn
x=arctan​−55−25​​​​+πn
Combine all the solutionsx=arctan​55+25​​​​+πn,x=arctan​−55+25​​​​+πn,x=arctan​55−25​​​​+πn,x=arctan​−55−25​​​​+πn
Show solutions in decimal formx=0.94247…+πn,x=−0.94247…+πn,x=0.31415…+πn,x=−0.31415…+πn

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