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Popular Trigonometry >

solvefor y,v=(1-4cos(5y))^{-1/2}

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Solution

solvefor

Solution

y=5arccos(−4v21−v2​)​+52πn​,y=−5arccos(−4v21−v2​)​+52πn​
Solution steps
v=(1−4cos(5y))−21​
Switch sides(1−4cos(5y))−21​=v
Take both sides of the equation to the power of −2:1−4cos(5y)=v21​
(1−4cos(5y))−21​=v
((1−4cos(5y))−21​)−2=v−2
Expand ((1−4cos(5y))−21​)−2:1−4cos(5y)
((1−4cos(5y))−21​)−2
Apply exponent rule: a−b=ab1​=((1−4cos(5y))−21​)21​
((1−4cos(5y))−21​)2:(1−4cos(5y))−1
Apply exponent rule: (ab)c=abc=(1−4cos(5y))(−21​)⋅2
(−21​)⋅2=−1
(−21​)⋅2
Remove parentheses: (−a)=−a=−21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=−21⋅2​
Cancel the common factor: 2=−1
=(1−4cos(5y))−1
=(1−4cos(5y))−11​
Apply exponent rule: a−1=a1​(1−4cos(5y))−1=(1−4cos(5y))1​=1−4cos(5y)1​1​
Apply the fraction rule: cb​1​=bc​=11−4cos(5y)​
Apply the fraction rule: 1a​=a=1−4cos(5y)
Refine=1−4cos(5y)
Expand v−2:v21​
v−2
Apply exponent rule: a−b=ab1​=v21​
1−4cos(5y)=v21​
1−4cos(5y)=v21​
Solve 1−4cos(5y)=v21​:cos(5y)=−4v21−v2​
1−4cos(5y)=v21​
Move 1to the right side
1−4cos(5y)=v21​
Subtract 1 from both sides1−4cos(5y)−1=v21​−1
Simplify−4cos(5y)=v21​−1
−4cos(5y)=v21​−1
Divide both sides by −4
−4cos(5y)=v21​−1
Divide both sides by −4−4−4cos(5y)​=−4v21​​−−41​
Simplify
−4−4cos(5y)​=−4v21​​−−41​
Simplify −4−4cos(5y)​:cos(5y)
−4−4cos(5y)​
Apply the fraction rule: −b−a​=ba​=44cos(5y)​
Divide the numbers: 44​=1=cos(5y)
Simplify −4v21​​−−41​:−4v21−v2​
−4v21​​−−41​
Apply rule ca​±cb​=ca±b​=−4v21​−1​
Apply the fraction rule: −ba​=−ba​=−4v21​−1​
Join v21​−1:v21−v2​
v21​−1
Convert element to fraction: 1=v21v2​=v21​−v21⋅v2​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=v21−1⋅v2​
Multiply: 1⋅v2=v2=v21−v2​
=−4v2−v2+1​​
Apply the fraction rule: acb​​=c⋅ab​=−4v2−v2+1​
cos(5y)=−4v21−v2​
cos(5y)=−4v21−v2​
cos(5y)=−4v21−v2​
cos(5y)=−4v21−v2​
Verify Solutions:cos(5y)=−4v21−v2​{v<0orv>0}
Check the solutions by plugging them into (1−4cos(5y))−21​=v
Remove the ones that don't agree with the equation.
Plugcos(5y)=−4v21−v2​:(1−4(−4v21−v2​))−21​=v⇒v<0orv>0
(1−4(−4v21−v2​))−21​=v
Take both sides of the equation to the power of −2:v21​=v21​
(1−4(−4v21−v2​))−21​=v
((1−4(−4v21−v2​))−21​)−2=v−2
Expand ((1−4(−4v21−v2​))−21​)−2:v21​
((1−4(−4v21−v2​))−21​)−2
Apply exponent rule: a−b=ab1​=((1−4(−4v21−v2​))−21​)21​
((1−4(−4v21−v2​))−21​)2:(1−4(−4v21−v2​))−1
Apply exponent rule: (ab)c=abc=(1−4(−4v21−v2​))(−21​)⋅2
(−21​)⋅2=−1
(−21​)⋅2
Remove parentheses: (−a)=−a=−21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=−21⋅2​
Cancel the common factor: 2=−1
=(1−4(−4v21−v2​))−1
=(1−4(−4v21−v2​))−11​
Apply exponent rule: a−1=a1​(1−4(−4v21−v2​))−1=(1−4(−4v21−v2​))1​=1−4(−4v21−v2​)1​1​
Apply the fraction rule: cb​1​=bc​=11−4(−4v21−v2​)​
Apply the fraction rule: 1a​=a=1−4(−4v21−v2​)
Expand 1−4(−4v21−v2​):v21​
(1−4(−4v21−v2​))
Apply rule −(−a)=a=1+4⋅4v21−v2​
4⋅4v21−v2​=v21−v2​
4⋅4v21−v2​
Multiply fractions: a⋅cb​=ca⋅b​=4v2(1−v2)⋅4​
Cancel the common factor: 4=v21−v2​
=1+v2−v2+1​
Apply the fraction rule: ca±b​=ca​±cb​v21−v2​=v21​−v2v2​=1+v21​−v2v2​
Apply rule aa​=1v2v2​=1=1+v21​−1
Group like terms=v21​+1−1
1−1=0=v21​
=v21​
Expand v−2:v21​
v−2
Apply exponent rule: a−b=ab1​=v21​
v21​=v21​
v21​=v21​
Solve v21​=v21​:True for all v
v21​=v21​
Subtract v21​ from both sidesv21​−v21​=v21​−v21​
Simplify0=0
Both sides are equal
Trueforallv
Verify Solutions
Find undefined (singularity) points:v=0
Take the denominator(s) of v21​ and compare to zero
Solve v2=0:v=0
v2=0
Apply rule xn=0⇒x=0
v=0
Take the denominator(s) of v21​ and compare to zero
Solve v2=0:v=0
v2=0
Apply rule xn=0⇒x=0
v=0
The following points are undefinedv=0
Since the equation is undefined for:0
Trueforallv
Trueforallv
Verify Solutions:v<0orv>0
(1−4(−4v21−v2​))−21​=v
Domain of (1−4(−4v21−v2​))−21​:v<0orv>0
Domain definition
Find undefined (singularity) points:v=0
(1−4(−4v21−v2​))−21​
Take the denominator(s) of (1−4(−4v21−v2​))−21​ and compare to zero
v=0
The following points are undefinedv=0
The function domainv<0orv>0
Combine domain interval with solution interval:Trueforallvand(v<0orv>0)
Merge Overlapping Intervalsv<0orv>0
The solution isv<0orv>0
The solution iscos(5y)=−4v21−v2​{v<0orv>0}
Apply trig inverse properties
cos(5y)=−4v21−v2​
General solutions for cos(5y)=−4v21−v2​cos(x)=a⇒x=arccos(a)+2πn,x=−arccos(a)+2πn5y=arccos(−4v21−v2​)+2πn,5y=−arccos(−4v21−v2​)+2πn
5y=arccos(−4v21−v2​)+2πn,5y=−arccos(−4v21−v2​)+2πn
Solve 5y=arccos(−4v21−v2​)+2πn:y=5arccos(−4v21−v2​)​+52πn​
5y=arccos(−4v21−v2​)+2πn
Divide both sides by 5
5y=arccos(−4v21−v2​)+2πn
Divide both sides by 555y​=5arccos(−4v21−v2​)​+52πn​
Simplifyy=5arccos(−4v21−v2​)​+52πn​
y=5arccos(−4v21−v2​)​+52πn​
Solve 5y=−arccos(−4v21−v2​)+2πn:y=−5arccos(−4v21−v2​)​+52πn​
5y=−arccos(−4v21−v2​)+2πn
Divide both sides by 5
5y=−arccos(−4v21−v2​)+2πn
Divide both sides by 555y​=−5arccos(−4v21−v2​)​+52πn​
Simplifyy=−5arccos(−4v21−v2​)​+52πn​
y=−5arccos(−4v21−v2​)​+52πn​
y=5arccos(−4v21−v2​)​+52πn​,y=−5arccos(−4v21−v2​)​+52πn​

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