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Popular Trigonometry >

2cos^2(4x+1)=0

  • Pre Algebra
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Solution

2cos2(4x+1)=0

Solution

x=−41​+8π​+2πn​,x=−41​+83π​+2πn​
+1
Degrees
x=8.17605…∘+90∘n,x=53.17605…∘+90∘n
Solution steps
2cos2(4x+1)=0
Divide both sides by 2
2cos2(4x+1)=0
Divide both sides by 2
2cos2(4x+1)=0
Divide both sides by 222cos2(4x+1)​=20​
Simplifycos2(4x+1)=0
cos2(4x+1)=0
Apply rule xn=0⇒x=0
cos(4x+1)=0
General solutions for cos(4x+1)=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​​​​
4x+1=2π​+2πn,4x+1=23π​+2πn
4x+1=2π​+2πn,4x+1=23π​+2πn
Solve 4x+1=2π​+2πn:x=−41​+8π​+2πn​
4x+1=2π​+2πn
Move 1to the right side
4x+1=2π​+2πn
Subtract 1 from both sides4x+1−1=2π​+2πn−1
Simplify4x=2π​+2πn−1
4x=2π​+2πn−1
Divide both sides by 4
4x=2π​+2πn−1
Divide both sides by 444x​=42π​​+42πn​−41​
Simplify
44x​=42π​​+42πn​−41​
Simplify 44x​:x
44x​
Divide the numbers: 44​=1=x
Simplify 42π​​+42πn​−41​:−41​+8π​+2πn​
42π​​+42πn​−41​
Group like terms=−41​+42πn​+42π​​
42πn​=2πn​
42πn​
Cancel the common factor: 2=2πn​
42π​​=8π​
42π​​
Apply the fraction rule: acb​​=c⋅ab​=2⋅4π​
Multiply the numbers: 2⋅4=8=8π​
=−41​+2πn​+8π​
Group like terms=−41​+8π​+2πn​
x=−41​+8π​+2πn​
x=−41​+8π​+2πn​
x=−41​+8π​+2πn​
Solve 4x+1=23π​+2πn:x=−41​+83π​+2πn​
4x+1=23π​+2πn
Move 1to the right side
4x+1=23π​+2πn
Subtract 1 from both sides4x+1−1=23π​+2πn−1
Simplify4x=23π​+2πn−1
4x=23π​+2πn−1
Divide both sides by 4
4x=23π​+2πn−1
Divide both sides by 444x​=423π​​+42πn​−41​
Simplify
44x​=423π​​+42πn​−41​
Simplify 44x​:x
44x​
Divide the numbers: 44​=1=x
Simplify 423π​​+42πn​−41​:−41​+83π​+2πn​
423π​​+42πn​−41​
Group like terms=−41​+42πn​+423π​​
42πn​=2πn​
42πn​
Cancel the common factor: 2=2πn​
423π​​=83π​
423π​​
Apply the fraction rule: acb​​=c⋅ab​=2⋅43π​
Multiply the numbers: 2⋅4=8=83π​
=−41​+2πn​+83π​
Group like terms=−41​+83π​+2πn​
x=−41​+83π​+2πn​
x=−41​+83π​+2πn​
x=−41​+83π​+2πn​
x=−41​+8π​+2πn​,x=−41​+83π​+2πn​

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Popular Examples

solvefor x,2sin(x)=-1solvefor t,f=acos((tpi)/6-(11pi)/(12))-9cos^2(θ)+9=16sin(θ)-7,0<= θ<3607cos(b)-sqrt(13)=0sin(x)=-1/3 ,0<= x<= 2pi

Frequently Asked Questions (FAQ)

  • What is the general solution for 2cos^2(4x+1)=0 ?

    The general solution for 2cos^2(4x+1)=0 is x=-1/4+pi/8+(pin)/2 ,x=-1/4+(3pi)/8+(pin)/2
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