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Popular Trigonometry >

1/(tan(x))-2tan(x)=-1/4

  • Pre Algebra
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Solution

tan(x)1​−2tan(x)=−41​

Solution

x=−0.57451…+πn,x=0.65766…+πn
+1
Degrees
x=−32.91754…∘+180∘n,x=37.68118…∘+180∘n
Solution steps
tan(x)1​−2tan(x)=−41​
Solve by substitution
tan(x)1​−2tan(x)=−41​
Let: tan(x)=uu1​−2u=−41​
u1​−2u=−41​:u=−16−1+129​​,u=161+129​​
u1​−2u=−41​
Multiply by LCM
u1​−2u=−41​
Find Least Common Multiplier of u,4:4u
u,4
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in u or 4=4u
Multiply by LCM=4uu1​⋅4u−2u⋅4u=−41​⋅4u
Simplify
u1​⋅4u−2u⋅4u=−41​⋅4u
Simplify u1​⋅4u:4
u1​⋅4u
Multiply fractions: a⋅cb​=ca⋅b​=u1⋅4u​
Cancel the common factor: u=1⋅4
Multiply the numbers: 1⋅4=4=4
Simplify −2u⋅4u:−8u2
−2u⋅4u
Multiply the numbers: 2⋅4=8=−8uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=−8u1+1
Add the numbers: 1+1=2=−8u2
Simplify −41​⋅4u:−u
−41​⋅4u
Multiply fractions: a⋅cb​=ca⋅b​=−41⋅4​u
Cancel the common factor: 4=−u⋅1
Multiply: u⋅1=u=−u
4−8u2=−u
4−8u2=−u
4−8u2=−u
Solve 4−8u2=−u:u=−16−1+129​​,u=161+129​​
4−8u2=−u
Move uto the left side
4−8u2=−u
Add u to both sides4−8u2+u=−u+u
Simplify4−8u2+u=0
4−8u2+u=0
Write in the standard form ax2+bx+c=0−8u2+u+4=0
Solve with the quadratic formula
−8u2+u+4=0
Quadratic Equation Formula:
For a=−8,b=1,c=4u1,2​=2(−8)−1±12−4(−8)⋅4​​
u1,2​=2(−8)−1±12−4(−8)⋅4​​
12−4(−8)⋅4​=129​
12−4(−8)⋅4​
Apply rule 1a=112=1=1−4(−8)⋅4​
Apply rule −(−a)=a=1+4⋅8⋅4​
Multiply the numbers: 4⋅8⋅4=128=1+128​
Add the numbers: 1+128=129=129​
u1,2​=2(−8)−1±129​​
Separate the solutionsu1​=2(−8)−1+129​​,u2​=2(−8)−1−129​​
u=2(−8)−1+129​​:−16−1+129​​
2(−8)−1+129​​
Remove parentheses: (−a)=−a=−2⋅8−1+129​​
Multiply the numbers: 2⋅8=16=−16−1+129​​
Apply the fraction rule: −ba​=−ba​=−16−1+129​​
u=2(−8)−1−129​​:161+129​​
2(−8)−1−129​​
Remove parentheses: (−a)=−a=−2⋅8−1−129​​
Multiply the numbers: 2⋅8=16=−16−1−129​​
Apply the fraction rule: −b−a​=ba​−1−129​=−(1+129​)=161+129​​
The solutions to the quadratic equation are:u=−16−1+129​​,u=161+129​​
u=−16−1+129​​,u=161+129​​
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of u1​−2u and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=−16−1+129​​,u=161+129​​
Substitute back u=tan(x)tan(x)=−16−1+129​​,tan(x)=161+129​​
tan(x)=−16−1+129​​,tan(x)=161+129​​
tan(x)=−16−1+129​​:x=arctan(−16−1+129​​)+πn
tan(x)=−16−1+129​​
Apply trig inverse properties
tan(x)=−16−1+129​​
General solutions for tan(x)=−16−1+129​​tan(x)=−a⇒x=arctan(−a)+πnx=arctan(−16−1+129​​)+πn
x=arctan(−16−1+129​​)+πn
tan(x)=161+129​​:x=arctan(161+129​​)+πn
tan(x)=161+129​​
Apply trig inverse properties
tan(x)=161+129​​
General solutions for tan(x)=161+129​​tan(x)=a⇒x=arctan(a)+πnx=arctan(161+129​​)+πn
x=arctan(161+129​​)+πn
Combine all the solutionsx=arctan(−16−1+129​​)+πn,x=arctan(161+129​​)+πn
Show solutions in decimal formx=−0.57451…+πn,x=0.65766…+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 1/(tan(x))-2tan(x)=-1/4 ?

    The general solution for 1/(tan(x))-2tan(x)=-1/4 is x=-0.57451…+pin,x=0.65766…+pin
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