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Popular Trigonometry >

n=(2tan^3(2θ)-1)^{1/2}

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Solution

n=(2tan3(2θ)−1)21​

Solution

Solution steps
n=(2tan3(2θ)−1)21​
Switch sides(2tan3(2θ)−1)21​=n
Square both sides:2tan3(2θ)−1=n2
(2tan3(2θ)−1)21​=n
((2tan3(2θ)−1)21​)2=n2
Expand ((2tan3(2θ)−1)21​)2:2tan3(2θ)−1
((2tan3(2θ)−1)21​)2
Apply exponent rule: (ab)c=abc=(2tan3(2θ)−1)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=2tan3(2θ)−1
2tan3(2θ)−1=n2
2tan3(2θ)−1=n2
Solve
2tan3(2θ)−1=n2
Move 1to the right side
2tan3(2θ)−1=n2
Add 1 to both sides2tan3(2θ)−1+1=n2+1
Simplify2tan3(2θ)=n2+1
2tan3(2θ)=n2+1
Divide both sides by 2
2tan3(2θ)=n2+1
Divide both sides by 222tan3(2θ)​=2n2​+21​
Simplify
22tan3(2θ)​=2n2​+21​
Simplify 22tan3(2θ)​:tan3(2θ)
22tan3(2θ)​
Divide the numbers: 22​=1=tan3(2θ)
Simplify 2n2​+21​:2n2+1​
2n2​+21​
Apply rule ca​±cb​=ca±b​=2n2+1​
tan3(2θ)=2n2+1​
tan3(2θ)=2n2+1​
tan3(2θ)=2n2+1​
For xn=f(a), n is odd, the solution is
Verify Solutions:
Check the solutions by plugging them into (2tan3(2θ)−1)21​=n
Remove the ones that don't agree with the equation.
Plug
Square both sides:n2=n2
Expand
Apply exponent rule: (ab)c=abc
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
Expand
Apply exponent rule: (ab)c=abc=(2n2+1​)31​⋅3
31​⋅3=1
31​⋅3
Multiply fractions: a⋅cb​=ca⋅b​=31⋅3​
Cancel the common factor: 3=1
=2n2+1​
=2⋅2n2+1​
Multiply fractions: a⋅cb​=ca⋅b​=2(n2+1)⋅2​
Cancel the common factor: 2=n2+1
=n2+1−1
1−1=0=n2
=n2
n2=n2
n2=n2
Both sides are equalTrueforalln
Verify Solutions:n<0False,n=0True,n>0True
Combine domain interval with solution interval:Trueforalln
Find the function intervals:n<0,n=0,n>0
Find the even roots arguments zeroes:
Solve
Factor
Rewrite as
Rewrite 2 as (231​)3
Rewrite 1 as 13
Apply exponent rule: ambm=(ab)m
Apply Difference of Cubes Formula: x3−y3=(x−y)(x2+xy+y2)
Refine
Apply exponent rule: (ab)c=abc=(2n2+1​)31​⋅2
31​⋅2=32​
31​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=31⋅2​
Multiply the numbers: 1⋅2=2=32​
=(2n2+1​)32​
Using the Zero Factor Principle: If ab=0then a=0or b=0
Solve
Move 1to the right side
Add 1 to both sides
Simplify
Divide both sides by 231​
Divide both sides by 231​
Simplify
Take both sides of the equation to the power of 3:2n2+1​=21​
Expand
Apply exponent rule: (ab)c=abc=(2n2+1​)31​⋅3
31​⋅3=1
31​⋅3
Multiply fractions: a⋅cb​=ca⋅b​=31⋅3​
Cancel the common factor: 3=1
=2n2+1​
Expand (231​1​)3:21​
(231​1​)3
Apply exponent rule: (ba​)c=bcac​=(231​)313​
(231​)3:2
Apply exponent rule: (ab)c=abc=231​⋅3
31​⋅3=1
31​⋅3
Multiply fractions: a⋅cb​=ca⋅b​=31⋅3​
Cancel the common factor: 3=1
=2
=213​
Apply rule 1a=113=1=21​
2n2+1​=21​
2n2+1​=21​
Solve 2n2+1​=21​:n=0
2n2+1​=21​
Multiply both sides by 2
2n2+1​=21​
Multiply both sides by 222(n2+1)​=21⋅2​
Simplify
22(n2+1)​=21⋅2​
Simplify 22(n2+1)​:n2+1
22(n2+1)​
Divide the numbers: 22​=1=n2+1
Simplify 21⋅2​:1
21⋅2​
Multiply the numbers: 1⋅2=2=22​
Apply rule aa​=1=1
n2+1=1
n2+1=1
n2+1=1
Move 1to the right side
n2+1=1
Subtract 1 from both sidesn2+1−1=1−1
Simplifyn2=0
n2=0
Apply rule xn=0⇒x=0
n=0
n=0
Verify Solutions:n=0True
Check the solutions by plugging them into
Remove the ones that don't agree with the equation.
Plug in n=0:True
Apply rule 0a=002=0
Add the numbers: 0+1=1
Apply exponent rule: ambm=(ab)m=(2⋅21​)31​
2⋅21​=1
2⋅21​
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=131​
Apply rule 1a=1=1
=1−1
Subtract the numbers: 1−1=0=0
0=0
True
The solution isn=0
Solve No Solution for n∈R
Use the following exponent property
Rewrite the equation with 232​u2+231​u+1=0
Solve 232​u2+231​u+1=0:No Solution for u∈R
232​u2+231​u+1=0
Discriminant 232​u2+231​u+1=0:−3⋅232​
232​u2+231​u+1=0
For a quadratic equation of the form ax2+bx+c=0 the discriminant is b2−4acFor a=232​,b=231​,c=1:(231​)2−4⋅232​⋅1(231​)2−4⋅232​⋅1
Expand (231​)2−4⋅232​⋅1:−3⋅232​
(231​)2−4⋅232​⋅1
(231​)2=232​
(231​)2
Apply exponent rule: (ab)c=abc=231​⋅2
31​⋅2=32​
31​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=31⋅2​
Multiply the numbers: 1⋅2=2=32​
=232​
4⋅232​⋅1=4⋅232​
4⋅232​⋅1
Multiply the numbers: 4⋅1=4=4⋅232​
=232​−4⋅232​
Add similar elements: 232​−4⋅232​=−3⋅232​=−3⋅232​
−3⋅232​
Discriminant cannot be negative for u∈R
The solution isNoSolutionforu∈R
NoSolutionforn∈R
n=0
Verify Solutions:n=0True
Check the solutions by plugging them into
Remove the ones that don't agree with the equation.
Plug in n=0:True
Apply rule 0a=002=0
Apply exponent rule: (ab)c=abc=(20+1​)31​⋅3
31​⋅3=1
31​⋅3
Multiply fractions: a⋅cb​=ca⋅b​=31⋅3​
Cancel the common factor: 3=1
=20+1​
Add the numbers: 0+1=1=21​
=2⋅21​
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1−1
Subtract the numbers: 1−1=0=0
0=0
True
The solution isn=0
n=0
The intervals are defined around the zeroes:n<0,n=0,n>0
Combine intervals with domainn<0,n=0,n>0
Check the solutions by plugging them into
Remove the ones that don't agree with the equation.
PlugFalse
The solution isn≥0
The solution is
Apply trig inverse properties
General solutions for tan(x)=a⇒x=arctan(a)+πk
Solve
Divide both sides by 2
Divide both sides by 2
Simplify

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Popular Examples

0=2sin(x+1)+1tan(θ)*10=104cos(x)+tan(45)=0.6sin(5x+8)=cos(9x-16)cos(x)=(sqrt(125))/(sqrt(174))

Frequently Asked Questions (FAQ)

  • What is the general solution for n=(2tan^3(2θ)-1)^{1/2} ?

    The general solution for n=(2tan^3(2θ)-1)^{1/2} is θ=(arctan(\sqrt[3]{(n^2+1)/2}))/(2)+(pik)/2
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