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Popular Trigonometry >

solvefor x,2cos(x)=2cos(3x)

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Solution

solvefor

Solution

x=2π​+2πn,x=23π​+2πn,x=π+2πn,x=2πn
+1
Degrees
x=90∘+360∘n,x=270∘+360∘n,x=180∘+360∘n,x=0∘+360∘n
Solution steps
2cos(x)=2cos(3x)
Subtract 2cos(3x) from both sides2cos(x)−2cos(3x)=0
Rewrite using trig identities
−2cos(3x)+2cos(x)
cos(3x)=4cos3(x)−3cos(x)
cos(3x)
Rewrite using trig identities
cos(3x)
Rewrite as=cos(2x+x)
Use the Angle Sum identity: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(2x)cos(x)−sin(2x)sin(x)
Use the Double Angle identity: sin(2x)=2sin(x)cos(x)=cos(2x)cos(x)−2sin(x)cos(x)sin(x)
Simplify cos(2x)cos(x)−2sin(x)cos(x)sin(x):cos(x)cos(2x)−2sin2(x)cos(x)
cos(2x)cos(x)−2sin(x)cos(x)sin(x)
2sin(x)cos(x)sin(x)=2sin2(x)cos(x)
2sin(x)cos(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=2cos(x)sin1+1(x)
Add the numbers: 1+1=2=2cos(x)sin2(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
Use the Double Angle identity: cos(2x)=2cos2(x)−1=(2cos2(x)−1)cos(x)−2sin2(x)cos(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
Expand (2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x):4cos3(x)−3cos(x)
(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
=cos(x)(2cos2(x)−1)−2cos(x)(1−cos2(x))
Expand cos(x)(2cos2(x)−1):2cos3(x)−cos(x)
cos(x)(2cos2(x)−1)
Apply the distributive law: a(b−c)=ab−aca=cos(x),b=2cos2(x),c=1=cos(x)2cos2(x)−cos(x)1
=2cos2(x)cos(x)−1cos(x)
Simplify 2cos2(x)cos(x)−1⋅cos(x):2cos3(x)−cos(x)
2cos2(x)cos(x)−1cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
Add the numbers: 2+1=3=2cos3(x)
1⋅cos(x)=cos(x)
1cos(x)
Multiply: 1⋅cos(x)=cos(x)=cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)−2(1−cos2(x))cos(x)
Expand −2cos(x)(1−cos2(x)):−2cos(x)+2cos3(x)
−2cos(x)(1−cos2(x))
Apply the distributive law: a(b−c)=ab−aca=−2cos(x),b=1,c=cos2(x)=−2cos(x)1−(−2cos(x))cos2(x)
Apply minus-plus rules−(−a)=a=−2⋅1cos(x)+2cos2(x)cos(x)
Simplify −2⋅1⋅cos(x)+2cos2(x)cos(x):−2cos(x)+2cos3(x)
−2⋅1cos(x)+2cos2(x)cos(x)
2⋅1⋅cos(x)=2cos(x)
2⋅1cos(x)
Multiply the numbers: 2⋅1=2=2cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
Add the numbers: 2+1=3=2cos3(x)
=−2cos(x)+2cos3(x)
=−2cos(x)+2cos3(x)
=2cos3(x)−cos(x)−2cos(x)+2cos3(x)
Simplify 2cos3(x)−cos(x)−2cos(x)+2cos3(x):4cos3(x)−3cos(x)
2cos3(x)−cos(x)−2cos(x)+2cos3(x)
Group like terms=2cos3(x)+2cos3(x)−cos(x)−2cos(x)
Add similar elements: 2cos3(x)+2cos3(x)=4cos3(x)=4cos3(x)−cos(x)−2cos(x)
Add similar elements: −cos(x)−2cos(x)=−3cos(x)=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=−2(4cos3(x)−3cos(x))+2cos(x)
Simplify −2(4cos3(x)−3cos(x))+2cos(x):−8cos3(x)+8cos(x)
−2(4cos3(x)−3cos(x))+2cos(x)
Expand −2(4cos3(x)−3cos(x)):−8cos3(x)+6cos(x)
−2(4cos3(x)−3cos(x))
Apply the distributive law: a(b−c)=ab−aca=−2,b=4cos3(x),c=3cos(x)=−2⋅4cos3(x)−(−2)⋅3cos(x)
Apply minus-plus rules−(−a)=a=−2⋅4cos3(x)+2⋅3cos(x)
Simplify −2⋅4cos3(x)+2⋅3cos(x):−8cos3(x)+6cos(x)
−2⋅4cos3(x)+2⋅3cos(x)
Multiply the numbers: 2⋅4=8=−8cos3(x)+2⋅3cos(x)
Multiply the numbers: 2⋅3=6=−8cos3(x)+6cos(x)
=−8cos3(x)+6cos(x)
=−8cos3(x)+6cos(x)+2cos(x)
Add similar elements: 6cos(x)+2cos(x)=8cos(x)=−8cos3(x)+8cos(x)
=−8cos3(x)+8cos(x)
8cos(x)−8cos3(x)=0
Solve by substitution
8cos(x)−8cos3(x)=0
Let: cos(x)=u8u−8u3=0
8u−8u3=0:u=0,u=−1,u=1
8u−8u3=0
Factor 8u−8u3:−8u(u+1)(u−1)
8u−8u3
Factor out common term −8u:−8u(u2−1)
−8u3+8u
Apply exponent rule: ab+c=abacu3=u2u=−8u2u+8u
Factor out common term −8u=−8u(u2−1)
=−8u(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)
=−8u(u+1)(u−1)
−8u(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
Substitute back u=cos(x)cos(x)=0,cos(x)=−1,cos(x)=1
cos(x)=0,cos(x)=−1,cos(x)=1
cos(x)=0:x=2π​+2πn,x=23π​+2πn
cos(x)=0
General solutions for cos(x)=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​​​​
x=2π​+2πn,x=23π​+2πn
x=2π​+2πn,x=23π​+2πn
cos(x)=−1:x=π+2πn
cos(x)=−1
General solutions for cos(x)=−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​​​​
x=π+2πn
x=π+2πn
cos(x)=1:x=2πn
cos(x)=1
General solutions for cos(x)=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​​​​
x=0+2πn
x=0+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn,x=π+2πn,x=2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for solvefor x,2cos(x)=2cos(3x) ?

    The general solution for solvefor x,2cos(x)=2cos(3x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=pi+2pin,x=2pin
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