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Popular Trigonometry >

2cos^2(x)-cos(x)=2-sec(x)

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Solution

2cos2(x)−cos(x)=2−sec(x)

Solution

x=3π​+2πn,x=35π​+2πn,x=π+2πn,x=2πn
+1
Degrees
x=60∘+360∘n,x=300∘+360∘n,x=180∘+360∘n,x=0∘+360∘n
Solution steps
2cos2(x)−cos(x)=2−sec(x)
Subtract 2−sec(x) from both sides2cos2(x)−cos(x)−2+sec(x)=0
Rewrite using trig identities
−2−cos(x)+sec(x)+2cos2(x)
Use the basic trigonometric identity: cos(x)=sec(x)1​=−2−sec(x)1​+sec(x)+2(sec(x)1​)2
2(sec(x)1​)2=sec2(x)2​
2(sec(x)1​)2
(sec(x)1​)2=sec2(x)1​
(sec(x)1​)2
Apply exponent rule: (ba​)c=bcac​=sec2(x)12​
Apply rule 1a=112=1=sec2(x)1​
=2⋅sec2(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=sec2(x)1⋅2​
Multiply the numbers: 1⋅2=2=sec2(x)2​
=−2−sec(x)1​+sec(x)+sec2(x)2​
−2−sec(x)1​+sec2(x)2​+sec(x)=0
Solve by substitution
−2−sec(x)1​+sec2(x)2​+sec(x)=0
Let: sec(x)=u−2−u1​+u22​+u=0
−2−u1​+u22​+u=0:u=2,u=−1,u=1
−2−u1​+u22​+u=0
Multiply by LCM
−2−u1​+u22​+u=0
Find Least Common Multiplier of u,u2:u2
u,u2
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in u or u2=u2
Multiply by LCM=u2−2u2−u1​u2+u22​u2+uu2=0⋅u2
Simplify
−2u2−u1​u2+u22​u2+uu2=0⋅u2
Simplify −u1​u2:−u
−u1​u2
Multiply fractions: a⋅cb​=ca⋅b​=−u1⋅u2​
Multiply: 1⋅u2=u2=−uu2​
Cancel the common factor: u=−u
Simplify u22​u2:2
u22​u2
Multiply fractions: a⋅cb​=ca⋅b​=u22u2​
Cancel the common factor: u2=2
Simplify uu2:u3
uu2
Apply exponent rule: ab⋅ac=ab+cuu2=u1+2=u1+2
Add the numbers: 1+2=3=u3
Simplify 0⋅u2:0
0⋅u2
Apply rule 0⋅a=0=0
−2u2−u+2+u3=0
−2u2−u+2+u3=0
−2u2−u+2+u3=0
Solve −2u2−u+2+u3=0:u=2,u=−1,u=1
−2u2−u+2+u3=0
Write in the standard form an​xn+…+a1​x+a0​=0u3−2u2−u+2=0
Factor u3−2u2−u+2:(u−2)(u+1)(u−1)
u3−2u2−u+2
=(u3−2u2)+(−u+2)
Factor out −1from −u+2:−(u−2)
−u+2
Factor out common term −1=−(u−2)
Factor out u2from u3−2u2:u2(u−2)
u3−2u2
Apply exponent rule: ab+c=abacu3=uu2=uu2−2u2
Factor out common term u2=u2(u−2)
=−(u−2)+u2(u−2)
Factor out common term u−2=(u−2)(u2−1)
Factor u2−1:(u+1)(u−1)
u2−1
Rewrite 1 as 12=u2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)u2−12=(u+1)(u−1)=(u+1)(u−1)
=(u−2)(u+1)(u−1)
(u−2)(u+1)(u−1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u−2=0oru+1=0oru−1=0
Solve u−2=0:u=2
u−2=0
Move 2to the right side
u−2=0
Add 2 to both sidesu−2+2=0+2
Simplifyu=2
u=2
Solve u+1=0:u=−1
u+1=0
Move 1to the right side
u+1=0
Subtract 1 from both sidesu+1−1=0−1
Simplifyu=−1
u=−1
Solve 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
The solutions areu=2,u=−1,u=1
u=2,u=−1,u=1
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −2−u1​+u22​+u and compare to zero
u=0
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=2,u=−1,u=1
Substitute back u=sec(x)sec(x)=2,sec(x)=−1,sec(x)=1
sec(x)=2,sec(x)=−1,sec(x)=1
sec(x)=2:x=3π​+2πn,x=35π​+2πn
sec(x)=2
General solutions for sec(x)=2
sec(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sec(x)1323​​2​2Undefined−2−2​−323​​​xπ67π​45π​34π​23π​35π​47π​611π​​sec(x)−1−323​​−2​−2Undefined22​323​​​​
x=3π​+2πn,x=35π​+2πn
x=3π​+2πn,x=35π​+2πn
sec(x)=−1:x=π+2πn
sec(x)=−1
General solutions for sec(x)=−1
sec(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sec(x)1323​​2​2Undefined−2−2​−323​​​xπ67π​45π​34π​23π​35π​47π​611π​​sec(x)−1−323​​−2​−2Undefined22​323​​​​
x=π+2πn
x=π+2πn
sec(x)=1:x=2πn
sec(x)=1
General solutions for sec(x)=1
sec(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sec(x)1323​​2​2Undefined−2−2​−323​​​xπ67π​45π​34π​23π​35π​47π​611π​​sec(x)−1−323​​−2​−2Undefined22​323​​​​
x=0+2πn
x=0+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
Combine all the solutionsx=3π​+2πn,x=35π​+2πn,x=π+2πn,x=2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 2cos^2(x)-cos(x)=2-sec(x) ?

    The general solution for 2cos^2(x)-cos(x)=2-sec(x) is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin,x=2pin
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