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Popular Trigonometry >

cos(3x)=cos(2x)cos(x)

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Solution

cos(3x)=cos(2x)cos(x)

Solution

x=2π​+2πn,x=23π​+2πn,x=2πn,x=π+2πn
+1
Degrees
x=90∘+360∘n,x=270∘+360∘n,x=0∘+360∘n,x=180∘+360∘n
Solution steps
cos(3x)=cos(2x)cos(x)
Subtract cos(2x)cos(x) from both sidescos(3x)−cos(2x)cos(x)=0
Rewrite using trig identities
cos(3x)−cos(2x)cos(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)
=4cos3(x)−3cos(x)−cos(2x)cos(x)
−3cos(x)+4cos3(x)−cos(2x)cos(x)=0
Factor −3cos(x)+4cos3(x)−cos(2x)cos(x):−cos(x)(3−4cos2(x)+cos(2x))
−3cos(x)+4cos3(x)−cos(2x)cos(x)
Apply exponent rule: ab+c=abaccos3(x)=cos(x)cos2(x)=−3cos(x)+4cos(x)cos2(x)−cos(x)cos(2x)
Factor out common term −cos(x)=−cos(x)(3−4cos2(x)+cos(2x))
−cos(x)(3−4cos2(x)+cos(2x))=0
Solving each part separatelycos(x)=0or3−4cos2(x)+cos(2x)=0
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
3−4cos2(x)+cos(2x)=0:x=2πn,x=π+2πn
3−4cos2(x)+cos(2x)=0
Rewrite using trig identities
3+cos(2x)−4cos2(x)
Use the Double Angle identity: cos(2x)=2cos2(x)−1=3+2cos2(x)−1−4cos2(x)
Simplify 3+2cos2(x)−1−4cos2(x):−2cos2(x)+2
3+2cos2(x)−1−4cos2(x)
Group like terms=2cos2(x)−4cos2(x)+3−1
Add similar elements: 2cos2(x)−4cos2(x)=−2cos2(x)=−2cos2(x)+3−1
Add/Subtract the numbers: 3−1=2=−2cos2(x)+2
=−2cos2(x)+2
2−2cos2(x)=0
Solve by substitution
2−2cos2(x)=0
Let: cos(x)=u2−2u2=0
2−2u2=0:u=1,u=−1
2−2u2=0
Move 2to the right side
2−2u2=0
Subtract 2 from both sides2−2u2−2=0−2
Simplify−2u2=−2
−2u2=−2
Divide both sides by −2
−2u2=−2
Divide both sides by −2−2−2u2​=−2−2​
Simplifyu2=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
Substitute back u=cos(x)cos(x)=1,cos(x)=−1
cos(x)=1,cos(x)=−1
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
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
Combine all the solutionsx=2πn,x=π+2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn,x=2πn,x=π+2πn

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Popular Examples

3cos^2(x)+sin^2(x)+5sin(x)=02tan(x)=5sin(x)5cot(x)=2tan(3x+(3pi)/4)=12sin(3x+1)=1

Frequently Asked Questions (FAQ)

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

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