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Popular Trigonometry >

2tan^2(x)+cot^2(x)-3=0

  • Pre Algebra
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Solution

2tan2(x)+cot2(x)−3=0

Solution

x=0.61547…+πn,x=2.52611…+πn,x=4π​+πn,x=43π​+πn
+1
Degrees
x=35.26438…∘+180∘n,x=144.73561…∘+180∘n,x=45∘+180∘n,x=135∘+180∘n
Solution steps
2tan2(x)+cot2(x)−3=0
Rewrite using trig identities
−3+cot2(x)+2tan2(x)
Use the basic trigonometric identity: tan(x)=cot(x)1​=−3+cot2(x)+2(cot(x)1​)2
2(cot(x)1​)2=cot2(x)2​
2(cot(x)1​)2
(cot(x)1​)2=cot2(x)1​
(cot(x)1​)2
Apply exponent rule: (ba​)c=bcac​=cot2(x)12​
Apply rule 1a=112=1=cot2(x)1​
=2⋅cot2(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=cot2(x)1⋅2​
Multiply the numbers: 1⋅2=2=cot2(x)2​
=−3+cot2(x)+cot2(x)2​
−3+cot2(x)+cot2(x)2​=0
Solve by substitution
−3+cot2(x)+cot2(x)2​=0
Let: cot(x)=u−3+u2+u22​=0
−3+u2+u22​=0:u=2​,u=−2​,u=1,u=−1
−3+u2+u22​=0
Multiply both sides by u2
−3+u2+u22​=0
Multiply both sides by u2−3u2+u2u2+u22​u2=0⋅u2
Simplify
−3u2+u2u2+u22​u2=0⋅u2
Simplify u2u2:u4
u2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=u2+2
Add the numbers: 2+2=4=u4
Simplify u22​u2:2
u22​u2
Multiply fractions: a⋅cb​=ca⋅b​=u22u2​
Cancel the common factor: u2=2
Simplify 0⋅u2:0
0⋅u2
Apply rule 0⋅a=0=0
−3u2+u4+2=0
−3u2+u4+2=0
−3u2+u4+2=0
Solve −3u2+u4+2=0:u=2​,u=−2​,u=1,u=−1
−3u2+u4+2=0
Write in the standard form an​xn+…+a1​x+a0​=0u4−3u2+2=0
Rewrite the equation with v=u2 and v2=u4v2−3v+2=0
Solve v2−3v+2=0:v=2,v=1
v2−3v+2=0
Solve with the quadratic formula
v2−3v+2=0
Quadratic Equation Formula:
For a=1,b=−3,c=2v1,2​=2⋅1−(−3)±(−3)2−4⋅1⋅2​​
v1,2​=2⋅1−(−3)±(−3)2−4⋅1⋅2​​
(−3)2−4⋅1⋅2​=1
(−3)2−4⋅1⋅2​
Apply exponent rule: (−a)n=an,if n is even(−3)2=32=32−4⋅1⋅2​
Multiply the numbers: 4⋅1⋅2=8=32−8​
32=9=9−8​
Subtract the numbers: 9−8=1=1​
Apply rule 1​=1=1
v1,2​=2⋅1−(−3)±1​
Separate the solutionsv1​=2⋅1−(−3)+1​,v2​=2⋅1−(−3)−1​
v=2⋅1−(−3)+1​:2
2⋅1−(−3)+1​
Apply rule −(−a)=a=2⋅13+1​
Add the numbers: 3+1=4=2⋅14​
Multiply the numbers: 2⋅1=2=24​
Divide the numbers: 24​=2=2
v=2⋅1−(−3)−1​:1
2⋅1−(−3)−1​
Apply rule −(−a)=a=2⋅13−1​
Subtract the numbers: 3−1=2=2⋅12​
Multiply the numbers: 2⋅1=2=22​
Apply rule aa​=1=1
The solutions to the quadratic equation are:v=2,v=1
v=2,v=1
Substitute back v=u2,solve for u
Solve u2=2:u=2​,u=−2​
u2=2
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=2​,u=−2​
Solve u2=1:u=1,u=−1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply rule 1​=1=1
−1​=−1
−1​
Apply rule 1​=1=−1
u=1,u=−1
The solutions are
u=2​,u=−2​,u=1,u=−1
u=2​,u=−2​,u=1,u=−1
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −3+u2+u22​ and compare to zero
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=2​,u=−2​,u=1,u=−1
Substitute back u=cot(x)cot(x)=2​,cot(x)=−2​,cot(x)=1,cot(x)=−1
cot(x)=2​,cot(x)=−2​,cot(x)=1,cot(x)=−1
cot(x)=2​:x=arccot(2​)+πn
cot(x)=2​
Apply trig inverse properties
cot(x)=2​
General solutions for cot(x)=2​cot(x)=a⇒x=arccot(a)+πnx=arccot(2​)+πn
x=arccot(2​)+πn
cot(x)=−2​:x=arccot(−2​)+πn
cot(x)=−2​
Apply trig inverse properties
cot(x)=−2​
General solutions for cot(x)=−2​cot(x)=−a⇒x=arccot(−a)+πnx=arccot(−2​)+πn
x=arccot(−2​)+πn
cot(x)=1:x=4π​+πn
cot(x)=1
General solutions for cot(x)=1
cot(x) periodicity table with πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cot(x)∓∞3​133​​0−33​​−1−3​​​
x=4π​+πn
x=4π​+πn
cot(x)=−1:x=43π​+πn
cot(x)=−1
General solutions for cot(x)=−1
cot(x) periodicity table with πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cot(x)∓∞3​133​​0−33​​−1−3​​​
x=43π​+πn
x=43π​+πn
Combine all the solutionsx=arccot(2​)+πn,x=arccot(−2​)+πn,x=4π​+πn,x=43π​+πn
Show solutions in decimal formx=0.61547…+πn,x=2.52611…+πn,x=4π​+πn,x=43π​+πn

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