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Popular Trigonometry >

cos(θ)cos(θ)=cot(θ)

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Solution

cos(θ)cos(θ)=cot(θ)

Solution

θ=2π​+2πn,θ=23π​+2πn
+1
Degrees
θ=90∘+360∘n,θ=270∘+360∘n
Solution steps
cos(θ)cos(θ)=cot(θ)
Subtract cot(θ) from both sidescos2(θ)−cot(θ)=0
Express with sin, cos
cos2(θ)−cot(θ)
Use the basic trigonometric identity: cot(x)=sin(x)cos(x)​=cos2(θ)−sin(θ)cos(θ)​
Simplify cos2(θ)−sin(θ)cos(θ)​:sin(θ)cos2(θ)sin(θ)−cos(θ)​
cos2(θ)−sin(θ)cos(θ)​
Convert element to fraction: cos2(θ)=sin(θ)cos2(θ)sin(θ)​=sin(θ)cos2(θ)sin(θ)​−sin(θ)cos(θ)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin(θ)cos2(θ)sin(θ)−cos(θ)​
=sin(θ)cos2(θ)sin(θ)−cos(θ)​
sin(θ)−cos(θ)+cos2(θ)sin(θ)​=0
g(x)f(x)​=0⇒f(x)=0−cos(θ)+cos2(θ)sin(θ)=0
Factor −cos(θ)+cos2(θ)sin(θ):cos(θ)(−1+sin(θ)cos(θ))
−cos(θ)+cos2(θ)sin(θ)
Apply exponent rule: ab+c=abacsin(θ)cos2(θ)=cos(θ)cos(θ)=−cos(θ)+cos(θ)cos(θ)
Factor out common term cos(θ)=cos(θ)(−1+sin(θ)cos(θ))
cos(θ)(−1+sin(θ)cos(θ))=0
Solving each part separatelycos(θ)=0or−1+sin(θ)cos(θ)=0
cos(θ)=0:θ=2π​+2πn,θ=23π​+2πn
cos(θ)=0
General solutions for cos(θ)=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​​​​
θ=2π​+2πn,θ=23π​+2πn
θ=2π​+2πn,θ=23π​+2πn
−1+sin(θ)cos(θ)=0:No Solution
−1+sin(θ)cos(θ)=0
Rewrite using trig identities
−1+sin(θ)cos(θ)
Use the Double Angle identity: 2sin(x)cos(x)=sin(2x)sin(x)cos(x)=2sin(2x)​=−1+2sin(2θ)​
−1+2sin(2θ)​=0
Move 1to the right side
−1+2sin(2θ)​=0
Add 1 to both sides−1+2sin(2θ)​+1=0+1
Simplify2sin(2θ)​=1
2sin(2θ)​=1
Multiply both sides by 2
2sin(2θ)​=1
Multiply both sides by 222sin(2θ)​=1⋅2
Simplifysin(2θ)=2
sin(2θ)=2
−1≤sin(x)≤1NoSolution
Combine all the solutionsθ=2π​+2πn,θ=23π​+2πn

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Popular Examples

3sin(θ)-cos(2θ)=1sin(θ)= 28/53 ,sin(2θ)arctan(x)=5tan(t)=5tan(t)=6

Frequently Asked Questions (FAQ)

  • What is the general solution for cos(θ)cos(θ)=cot(θ) ?

    The general solution for cos(θ)cos(θ)=cot(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin
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