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Popular Trigonometry >

((cos^3(a)))/((2cos^2(a)-1))=cos(a)

  • Pre Algebra
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Solution

(2cos2(a)−1)(cos3(a))​=cos(a)

Solution

a=2π​+2πn,a=23π​+2πn,a=π+2πn,a=2πn
+1
Degrees
a=90∘+360∘n,a=270∘+360∘n,a=180∘+360∘n,a=0∘+360∘n
Solution steps
(2cos2(a)−1)(cos3(a))​=cos(a)
Solve by substitution
2cos2(a)−1cos3(a)​=cos(a)
Let: cos(a)=u2u2−1u3​=u
2u2−1u3​=u:u=0,u=−1,u=1
2u2−1u3​=u
Multiply both sides by 2u2−1
2u2−1u3​=u
Multiply both sides by 2u2−12u2−1u3​(2u2−1)=u(2u2−1)
Simplifyu3=u(2u2−1)
u3=u(2u2−1)
Solve u3=u(2u2−1):u=0,u=−1,u=1
u3=u(2u2−1)
Expand u(2u2−1):2u3−u
u(2u2−1)
Apply the distributive law: a(b−c)=ab−aca=u,b=2u2,c=1=u⋅2u2−u⋅1
=2u2u−1⋅u
Simplify 2u2u−1⋅u:2u3−u
2u2u−1⋅u
2u2u=2u3
2u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=2u2+1
Add the numbers: 2+1=3=2u3
1⋅u=u
1⋅u
Multiply: 1⋅u=u=u
=2u3−u
=2u3−u
u3=2u3−u
Switch sides2u3−u=u3
Move u3to the left side
2u3−u=u3
Subtract u3 from both sides2u3−u−u3=u3−u3
Simplifyu3−u=0
u3−u=0
Factor u3−u:u(u+1)(u−1)
u3−u
Factor out common term u:u(u2−1)
u3−u
Apply exponent rule: ab+c=abacu3=u2u=u2u−u
Factor out common term u=u(u2−1)
=u(u2−1)
Factor u2−1:(u+1)(u−1)
u2−1
Rewrite 1 as 12=u2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)u2−12=(u+1)(u−1)=(u+1)(u−1)
=u(u+1)(u−1)
u(u+1)(u−1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u=0oru+1=0oru−1=0
Solve u+1=0:u=−1
u+1=0
Move 1to the right side
u+1=0
Subtract 1 from both sidesu+1−1=0−1
Simplifyu=−1
u=−1
Solve u−1=0:u=1
u−1=0
Move 1to the right side
u−1=0
Add 1 to both sidesu−1+1=0+1
Simplifyu=1
u=1
The solutions areu=0,u=−1,u=1
u=0,u=−1,u=1
Verify Solutions
Find undefined (singularity) points:u=2​1​,u=−2​1​
Take the denominator(s) of 2u2−1u3​ and compare to zero
Solve 2u2−1=0:u=2​1​,u=−2​1​
2u2−1=0
Move 1to the right side
2u2−1=0
Add 1 to both sides2u2−1+1=0+1
Simplify2u2=1
2u2=1
Divide both sides by 2
2u2=1
Divide both sides by 222u2​=21​
Simplifyu2=21​
u2=21​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=21​​,u=−21​​
21​​=2​1​
21​​
Apply radical rule: ba​​=b​a​​,a≥0,b≥0=2​1​​
Apply radical rule: 1​=11​=1=2​1​
−21​​=−2​1​
−21​​
Apply radical rule: ba​​=b​a​​,a≥0,b≥0=−2​1​​
Apply radical rule: 1​=11​=1=−2​1​
u=2​1​,u=−2​1​
The following points are undefinedu=2​1​,u=−2​1​
Combine undefined points with solutions:
u=0,u=−1,u=1
Substitute back u=cos(a)cos(a)=0,cos(a)=−1,cos(a)=1
cos(a)=0,cos(a)=−1,cos(a)=1
cos(a)=0:a=2π​+2πn,a=23π​+2πn
cos(a)=0
General solutions for cos(a)=0
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
a=2π​+2πn,a=23π​+2πn
a=2π​+2πn,a=23π​+2πn
cos(a)=−1:a=π+2πn
cos(a)=−1
General solutions for cos(a)=−1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
a=π+2πn
a=π+2πn
cos(a)=1:a=2πn
cos(a)=1
General solutions for cos(a)=1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
a=0+2πn
a=0+2πn
Solve a=0+2πn:a=2πn
a=0+2πn
0+2πn=2πna=2πn
a=2πn
Combine all the solutionsa=2π​+2πn,a=23π​+2πn,a=π+2πn,a=2πn

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