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Popular Trigonometry >

cos(1/(3x))= 1/3

  • Pre Algebra
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Solution

cos(3x1​)=31​

Solution

x=3(1.23095…+2πn)1​,x=3(2π−1.23095…+2πn)1​
+1
Degrees
x=0∘+2.54168…∘n,x=0∘+1.68486…∘n
Solution steps
cos(3x1​)=31​
Apply trig inverse properties
cos(3x1​)=31​
General solutions for cos(3x1​)=31​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πn3x1​=arccos(31​)+2πn,3x1​=2π−arccos(31​)+2πn
3x1​=arccos(31​)+2πn,3x1​=2π−arccos(31​)+2πn
Solve 3x1​=arccos(31​)+2πn:x=3(arccos(31​)+2πn)1​;n=−2πarccos(31​)​
3x1​=arccos(31​)+2πn
Multiply both sides by x
3x1​=arccos(31​)+2πn
Multiply both sides by x3x1​x=arccos(31​)x+2πnx
Simplify31​=arccos(31​)x+2πnx
31​=arccos(31​)x+2πnx
Switch sidesarccos(31​)x+2πnx=31​
Factor arccos(31​)x+2πnx:x(arccos(31​)+2πn)
arccos(31​)x+2πnx
Factor out common term x=x(arccos(31​)+2πn)
x(arccos(31​)+2πn)=31​
Divide both sides by arccos(31​)+2πn;n=−2πarccos(31​)​
x(arccos(31​)+2πn)=31​
Divide both sides by arccos(31​)+2πn;n=−2πarccos(31​)​arccos(31​)+2πnx(arccos(31​)+2πn)​=arccos(31​)+2πn31​​;n=−2πarccos(31​)​
Simplify
arccos(31​)+2πnx(arccos(31​)+2πn)​=arccos(31​)+2πn31​​
Simplify arccos(31​)+2πnx(arccos(31​)+2πn)​:x
arccos(31​)+2πnx(arccos(31​)+2πn)​
Cancel the common factor: arccos(31​)+2πn=x
Simplify arccos(31​)+2πn31​​:3(arccos(31​)+2πn)1​
arccos(31​)+2πn31​​
Apply the fraction rule: acb​​=c⋅ab​=3(arccos(31​)+2πn)1​
x=3(arccos(31​)+2πn)1​;n=−2πarccos(31​)​
x=3(arccos(31​)+2πn)1​;n=−2πarccos(31​)​
x=3(arccos(31​)+2πn)1​;n=−2πarccos(31​)​
Solve 3x1​=2π−arccos(31​)+2πn:x=3(2π−arccos(31​)+2πn)1​;n=2π−2π+arccos(31​)​
3x1​=2π−arccos(31​)+2πn
Multiply both sides by x
3x1​=2π−arccos(31​)+2πn
Multiply both sides by x3x1​x=2πx−arccos(31​)x+2πnx
Simplify31​=2πx−arccos(31​)x+2πnx
31​=2πx−arccos(31​)x+2πnx
Switch sides2πx−arccos(31​)x+2πnx=31​
Factor 2πx−arccos(31​)x+2πnx:x(2π−arccos(31​)+2πn)
2πx−arccos(31​)x+2πnx
Factor out common term x=x(2π−arccos(31​)+2πn)
x(2π−arccos(31​)+2πn)=31​
Divide both sides by 2π−arccos(31​)+2πn;n=2π−2π+arccos(31​)​
x(2π−arccos(31​)+2πn)=31​
Divide both sides by 2π−arccos(31​)+2πn;n=2π−2π+arccos(31​)​2π−arccos(31​)+2πnx(2π−arccos(31​)+2πn)​=2π−arccos(31​)+2πn31​​;n=2π−2π+arccos(31​)​
Simplify
2π−arccos(31​)+2πnx(2π−arccos(31​)+2πn)​=2π−arccos(31​)+2πn31​​
Simplify 2π−arccos(31​)+2πnx(2π−arccos(31​)+2πn)​:x
2π−arccos(31​)+2πnx(2π−arccos(31​)+2πn)​
Cancel the common factor: 2π−arccos(31​)+2πn=x
Simplify 2π−arccos(31​)+2πn31​​:3(2π−arccos(31​)+2πn)1​
2π−arccos(31​)+2πn31​​
Apply the fraction rule: acb​​=c⋅ab​=3(2π−arccos(31​)+2πn)1​
x=3(2π−arccos(31​)+2πn)1​;n=2π−2π+arccos(31​)​
x=3(2π−arccos(31​)+2πn)1​;n=2π−2π+arccos(31​)​
x=3(2π−arccos(31​)+2πn)1​;n=2π−2π+arccos(31​)​
x=3(arccos(31​)+2πn)1​,x=3(2π−arccos(31​)+2πn)1​
Show solutions in decimal formx=3(1.23095…+2πn)1​,x=3(2π−1.23095…+2πn)1​

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Popular Examples

arctan(1+x)+arctan(1-x)=arctan(1/2)-4cos^2(x)=0sin^2(x)-4sin^2(x)+7cos^2(x)=0(sin(x))(cos(x))=0sin^2(x)-15sin(x)cos(x)+50cos^2(x)=0

Frequently Asked Questions (FAQ)

  • What is the general solution for cos(1/(3x))= 1/3 ?

    The general solution for cos(1/(3x))= 1/3 is x= 1/(3(1.23095…+2pin)),x= 1/(3(2pi-1.23095…+2pin))
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