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Popular Trigonometry >

2cos^2(x)-cos(x)-1<0

  • Pre Algebra
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Solution

2cos2(x)−cos(x)−1<0

Solution

2πn<x<32π​+2πnor34π​+2πn<x<2π+2πn
+2
Interval Notation
(2πn,32π​+2πn)∪(34π​+2πn,2π+2πn)
Decimal
2πn<x<2.09439…+2πnor4.18879…+2πn<x<6.28318…+2πn
Solution steps
2cos2(x)−cos(x)−1<0
Let: u=cos(x)2u2−u−1<0
2u2−u−1<0:−21​<u<1
2u2−u−1<0
Factor 2u2−u−1:(2u+1)(u−1)
2u2−u−1
Break the expression into groups
2u2−u−1
Definition
Factors of 2:1,2
2
Divisors (Factors)
Find the Prime factors of 2:2
2
2 is a prime number, therefore no factorization is possible=2
Add 1 1
The factors of 21,2
Negative factors of 2:−1,−2
Multiply the factors by −1 to get the negative factors−1,−2
For every two factors such that u∗v=−2,check if u+v=−1
Check u=1,v=−2:u∗v=−2,u+v=−1⇒TrueCheck u=2,v=−1:u∗v=−2,u+v=1⇒False
u=1,v=−2
Group into (ax2+ux)+(vx+c)(2u2+u)+(−2u−1)
=(2u2+u)+(−2u−1)
Factor out ufrom 2u2+u:u(2u+1)
2u2+u
Apply exponent rule: ab+c=abacu2=uu=2uu+u
Factor out common term u=u(2u+1)
Factor out −1from −2u−1:−(2u+1)
−2u−1
Factor out common term −1=−(2u+1)
=u(2u+1)−(2u+1)
Factor out common term 2u+1=(2u+1)(u−1)
(2u+1)(u−1)<0
Identify the intervals
Find the signs of the factors of (2u+1)(u−1)
Find the signs of 2u+1
2u+1=0:u=−21​
2u+1=0
Move 1to the right side
2u+1=0
Subtract 1 from both sides2u+1−1=0−1
Simplify2u=−1
2u=−1
Divide both sides by 2
2u=−1
Divide both sides by 222u​=2−1​
Simplifyu=−21​
u=−21​
2u+1<0:u<−21​
2u+1<0
Move 1to the right side
2u+1<0
Subtract 1 from both sides2u+1−1<0−1
Simplify2u<−1
2u<−1
Divide both sides by 2
2u<−1
Divide both sides by 222u​<2−1​
Simplifyu<−21​
u<−21​
2u+1>0:u>−21​
2u+1>0
Move 1to the right side
2u+1>0
Subtract 1 from both sides2u+1−1>0−1
Simplify2u>−1
2u>−1
Divide both sides by 2
2u>−1
Divide both sides by 222u​>2−1​
Simplifyu>−21​
u>−21​
Find the signs of u−1
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
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
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
Summarize in a table:2u+1u−1(2u+1)(u−1)​u<−21​−−+​u=−21​0−0​−21​<u<1+−−​u=1+00​u>1+++​​
Identify the intervals that satisfy the required condition: <0−21​<u<1
−21​<u<1
−21​<u<1
Substitute back u=cos(x)−21​<cos(x)<1
If a<u<bthen a<uandu<b−21​<cos(x)andcos(x)<1
−21​<cos(x):−32π​+2πn<x<32π​+2πn
−21​<cos(x)
Switch sidescos(x)>−21​
For cos(x)>a, if −1≤a<1 then −arccos(a)+2πn<x<arccos(a)+2πn−arccos(−21​)+2πn<x<arccos(−21​)+2πn
Simplify −arccos(−21​):−32π​
−arccos(−21​)
Use the following trivial identity:arccos(−21​)=32π​x−1−23​​−22​​−21​021​22​​23​​1​arccos(x)π65π​43π​32π​2π​3π​4π​6π​0​arccos(x)180∘150∘135∘120∘90∘60∘45∘30∘0∘​​=−32π​
Simplify arccos(−21​):32π​
arccos(−21​)
Use the following trivial identity:arccos(−21​)=32π​x−1−23​​−22​​−21​021​22​​23​​1​arccos(x)π65π​43π​32π​2π​3π​4π​6π​0​arccos(x)180∘150∘135∘120∘90∘60∘45∘30∘0∘​​=32π​
−32π​+2πn<x<32π​+2πn
cos(x)<1:2πn<x<2π+2πn
cos(x)<1
For cos(x)<a, if −1<a≤1 then arccos(a)+2πn<x<2π−arccos(a)+2πnarccos(1)+2πn<x<2π−arccos(1)+2πn
Simplify arccos(1):0
arccos(1)
Use the following trivial identity:arccos(1)=0x−1−23​​−22​​−21​021​22​​23​​1​arccos(x)π65π​43π​32π​2π​3π​4π​6π​0​arccos(x)180∘150∘135∘120∘90∘60∘45∘30∘0∘​​=0
Simplify 2π−arccos(1):2π
2π−arccos(1)
Use the following trivial identity:arccos(1)=0x−1−23​​−22​​−21​021​22​​23​​1​arccos(x)π65π​43π​32π​2π​3π​4π​6π​0​arccos(x)180∘150∘135∘120∘90∘60∘45∘30∘0∘​​=2π−0
2π−0=2π=2π
0+2πn<x<2π+2πn
Simplify2πn<x<2π+2πn
Combine the intervals−32π​+2πn<x<32π​+2πnand2πn<x<2π+2πn
Merge Overlapping Intervals2πn<x<32π​+2πnor34π​+2πn<x<2π+2πn

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