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Popular Trigonometry >

tan^2(x)-2tan(x)=1

  • Pre Algebra
  • Algebra
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Solution

tan2(x)−2tan(x)=1

Solution

x=1.17809…+πn,x=−0.39269…+πn
+1
Degrees
x=67.5∘+180∘n,x=−22.5∘+180∘n
Solution steps
tan2(x)−2tan(x)=1
Solve by substitution
tan2(x)−2tan(x)=1
Let: tan(x)=uu2−2u=1
u2−2u=1:u=1+2​,u=1−2​
u2−2u=1
Move 1to the left side
u2−2u=1
Subtract 1 from both sidesu2−2u−1=1−1
Simplifyu2−2u−1=0
u2−2u−1=0
Solve with the quadratic formula
u2−2u−1=0
Quadratic Equation Formula:
For a=1,b=−2,c=−1u1,2​=2⋅1−(−2)±(−2)2−4⋅1⋅(−1)​​
u1,2​=2⋅1−(−2)±(−2)2−4⋅1⋅(−1)​​
(−2)2−4⋅1⋅(−1)​=22​
(−2)2−4⋅1⋅(−1)​
Apply rule −(−a)=a=(−2)2+4⋅1⋅1​
Apply exponent rule: (−a)n=an,if n is even(−2)2=22=22+4⋅1⋅1​
Multiply the numbers: 4⋅1⋅1=4=22+4​
22=4=4+4​
Add the numbers: 4+4=8=8​
Prime factorization of 8:23
8
8divides by 28=4⋅2=2⋅4
4divides by 24=2⋅2=2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2
=23
=23​
Apply exponent rule: ab+c=ab⋅ac=22⋅2​
Apply radical rule: =2​22​
Apply radical rule: 22​=2=22​
u1,2​=2⋅1−(−2)±22​​
Separate the solutionsu1​=2⋅1−(−2)+22​​,u2​=2⋅1−(−2)−22​​
u=2⋅1−(−2)+22​​:1+2​
2⋅1−(−2)+22​​
Apply rule −(−a)=a=2⋅12+22​​
Multiply the numbers: 2⋅1=2=22+22​​
Factor 2+22​:2(1+2​)
2+22​
Rewrite as=2⋅1+22​
Factor out common term 2=2(1+2​)
=22(1+2​)​
Divide the numbers: 22​=1=1+2​
u=2⋅1−(−2)−22​​:1−2​
2⋅1−(−2)−22​​
Apply rule −(−a)=a=2⋅12−22​​
Multiply the numbers: 2⋅1=2=22−22​​
Factor 2−22​:2(1−2​)
2−22​
Rewrite as=2⋅1−22​
Factor out common term 2=2(1−2​)
=22(1−2​)​
Divide the numbers: 22​=1=1−2​
The solutions to the quadratic equation are:u=1+2​,u=1−2​
Substitute back u=tan(x)tan(x)=1+2​,tan(x)=1−2​
tan(x)=1+2​,tan(x)=1−2​
tan(x)=1+2​:x=arctan(1+2​)+πn
tan(x)=1+2​
Apply trig inverse properties
tan(x)=1+2​
General solutions for tan(x)=1+2​tan(x)=a⇒x=arctan(a)+πnx=arctan(1+2​)+πn
x=arctan(1+2​)+πn
tan(x)=1−2​:x=arctan(1−2​)+πn
tan(x)=1−2​
Apply trig inverse properties
tan(x)=1−2​
General solutions for tan(x)=1−2​tan(x)=−a⇒x=arctan(−a)+πnx=arctan(1−2​)+πn
x=arctan(1−2​)+πn
Combine all the solutionsx=arctan(1+2​)+πn,x=arctan(1−2​)+πn
Show solutions in decimal formx=1.17809…+πn,x=−0.39269…+πn

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sin(a)=0.253tan^2(x+15)-1=03sin^4(x)+cos^4(x)=1sec(3x)=53sin^2(x)+2sin(x)cos^2(x/2)-sin(x)=0

Frequently Asked Questions (FAQ)

  • What is the general solution for tan^2(x)-2tan(x)=1 ?

    The general solution for tan^2(x)-2tan(x)=1 is x=1.17809…+pin,x=-0.39269…+pin
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