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Popular Trigonometry >

cos(2x)=cos(x)-cos(30)[180]

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Solution

cos(2x)=cos(x)−cos(30∘)[180]

Solution

NoSolutionforx∈R
Solution steps
cos(2x)=cos(x)−cos(30∘)[180]
cos(30∘)=23​​
cos(30∘)
Use the following trivial identity:cos(30∘)=23​​
cos(30∘)
cos(x) periodicity table with 360∘n cycle:
x030∘45∘60∘90∘120∘135∘150∘​cos(x)123​​22​​21​0−21​−22​​−23​​​x180∘210∘225∘240∘270∘300∘315∘330∘​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
=23​​
=23​​
cos(2x)=cos(x)−23​​[180]
Subtract cos(x)−23​​[180] from both sidescos(2x)−cos(x)+903​=0
Rewrite using trig identities
cos(2x)−cos(x)+903​
Use the Double Angle identity: cos(2x)=2cos2(x)−1=2cos2(x)−1−cos(x)+903​
−1−cos(x)+2cos2(x)+903​=0
Solve by substitution
−1−cos(x)+2cos2(x)+903​=0
Let: cos(x)=u−1−u+2u2+903​=0
−1−u+2u2+903​=0:u=41​+i43−1+803​​​,u=41​−i43−1+803​​​
−1−u+2u2+903​=0
Write in the standard form ax2+bx+c=02u2−u−1+903​=0
Solve with the quadratic formula
2u2−u−1+903​=0
Quadratic Equation Formula:
For a=2,b=−1,c=−1+903​u1,2​=2⋅2−(−1)±(−1)2−4⋅2(−1+903​)​​
u1,2​=2⋅2−(−1)±(−1)2−4⋅2(−1+903​)​​
Simplify (−1)2−4⋅2(−1+903​)​:3i803​−1​
(−1)2−4⋅2(−1+903​)​
(−1)2=1
(−1)2
Apply exponent rule: (−a)n=an,if n is even(−1)2=12=12
Apply rule 1a=1=1
4⋅2(−1+903​)=8(−1+903​)
4⋅2(−1+903​)
Multiply the numbers: 4⋅2=8=8(903​−1)
=1−8(903​−1)​
Apply imaginary number rule: −a​=ia​=i8(903​−1)−1​
−1+8(−1+903​)​=3803​−1​
−1+8(−1+903​)​
Expand −1+8(−1+903​):7203​−9
−1+8(−1+903​)
Expand 8(−1+903​):−8+7203​
8(−1+903​)
Apply the distributive law: a(b+c)=ab+aca=8,b=−1,c=903​=8(−1)+8⋅903​
Apply minus-plus rules+(−a)=−a=−8⋅1+8⋅903​
Simplify −8⋅1+8⋅903​:−8+7203​
−8⋅1+8⋅903​
Multiply the numbers: 8⋅1=8=−8+8⋅903​
Multiply the numbers: 8⋅90=720=−8+7203​
=−8+7203​
=−1−8+7203​
Subtract the numbers: −1−8=−9=7203​−9
=7203​−9​
Factor 7203​−9:9(803​−1)
7203​−9
Rewrite as=9⋅803​−9⋅1
Factor out common term 9=9(803​−1)
=9(803​−1)​
Apply radical rule: assuming a≥0,b≥0=9​803​−1​
9​=3
9​
Factor the number: 9=32=32​
Apply radical rule: 32​=3=3
=3803​−1​
=3i803​−1​
u1,2​=2⋅2−(−1)±3i803​−1​​
Separate the solutionsu1​=2⋅2−(−1)+3i803​−1​​,u2​=2⋅2−(−1)−3i803​−1​​
u=2⋅2−(−1)+3i803​−1​​:41​+i43−1+803​​​
2⋅2−(−1)+3i803​−1​​
Apply rule −(−a)=a=2⋅21+3i803​−1​​
Multiply the numbers: 2⋅2=4=41+3i803​−1​​
Rewrite 41+3i803​−1​​ in standard complex form: 41​+43803​−1​​i
41+3i803​−1​​
Apply the fraction rule: ca±b​=ca​±cb​41+3i803​−1​​=41​+43i803​−1​​=41​+43i803​−1​​
=41​+43803​−1​​i
u=2⋅2−(−1)−3i803​−1​​:41​−i43−1+803​​​
2⋅2−(−1)−3i803​−1​​
Apply rule −(−a)=a=2⋅21−3i803​−1​​
Multiply the numbers: 2⋅2=4=41−3i803​−1​​
Rewrite 41−3i803​−1​​ in standard complex form: 41​−43803​−1​​i
41−3i803​−1​​
Apply the fraction rule: ca±b​=ca​±cb​41−3i803​−1​​=41​−43i803​−1​​=41​−43i803​−1​​
=41​−43803​−1​​i
The solutions to the quadratic equation are:u=41​+i43−1+803​​​,u=41​−i43−1+803​​​
Substitute back u=cos(x)cos(x)=41​+i43−1+803​​​,cos(x)=41​−i43−1+803​​​
cos(x)=41​+i43−1+803​​​,cos(x)=41​−i43−1+803​​​
cos(x)=41​+i43−1+803​​​:No Solution
cos(x)=41​+i43−1+803​​​
NoSolution
cos(x)=41​−i43−1+803​​​:No Solution
cos(x)=41​−i43−1+803​​​
NoSolution
Combine all the solutionsNoSolutionforx∈R

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Frequently Asked Questions (FAQ)

  • What is the general solution for cos(2x)=cos(x)-cos(30)[180] ?

    The general solution for cos(2x)=cos(x)-cos(30)[180] is No Solution for x\in\mathbb{R}
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