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Popular Trigonometry >

2cot^2(3x)=5csc(3x)-4

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Solution

2cot2(3x)=5csc(3x)−4

Solution

x=18π​+32πn​,x=185π​+32πn​
+1
Degrees
x=10∘+120∘n,x=50∘+120∘n
Solution steps
2cot2(3x)=5csc(3x)−4
Subtract 5csc(3x)−4 from both sides2cot2(3x)−5csc(3x)+4=0
Rewrite using trig identities
4+2cot2(3x)−5csc(3x)
Use the Pythagorean identity: 1+cot2(x)=csc2(x)cot2(x)=csc2(x)−1=4+2(csc2(3x)−1)−5csc(3x)
Simplify 4+2(csc2(3x)−1)−5csc(3x):2csc2(3x)−5csc(3x)+2
4+2(csc2(3x)−1)−5csc(3x)
Expand 2(csc2(3x)−1):2csc2(3x)−2
2(csc2(3x)−1)
Apply the distributive law: a(b−c)=ab−aca=2,b=csc2(3x),c=1=2csc2(3x)−2⋅1
Multiply the numbers: 2⋅1=2=2csc2(3x)−2
=4+2csc2(3x)−2−5csc(3x)
Simplify 4+2csc2(3x)−2−5csc(3x):2csc2(3x)−5csc(3x)+2
4+2csc2(3x)−2−5csc(3x)
Group like terms=2csc2(3x)−5csc(3x)+4−2
Add/Subtract the numbers: 4−2=2=2csc2(3x)−5csc(3x)+2
=2csc2(3x)−5csc(3x)+2
=2csc2(3x)−5csc(3x)+2
2+2csc2(3x)−5csc(3x)=0
Solve by substitution
2+2csc2(3x)−5csc(3x)=0
Let: csc(3x)=u2+2u2−5u=0
2+2u2−5u=0:u=2,u=21​
2+2u2−5u=0
Write in the standard form ax2+bx+c=02u2−5u+2=0
Solve with the quadratic formula
2u2−5u+2=0
Quadratic Equation Formula:
For a=2,b=−5,c=2u1,2​=2⋅2−(−5)±(−5)2−4⋅2⋅2​​
u1,2​=2⋅2−(−5)±(−5)2−4⋅2⋅2​​
(−5)2−4⋅2⋅2​=3
(−5)2−4⋅2⋅2​
Apply exponent rule: (−a)n=an,if n is even(−5)2=52=52−4⋅2⋅2​
Multiply the numbers: 4⋅2⋅2=16=52−16​
52=25=25−16​
Subtract the numbers: 25−16=9=9​
Factor the number: 9=32=32​
Apply radical rule: 32​=3=3
u1,2​=2⋅2−(−5)±3​
Separate the solutionsu1​=2⋅2−(−5)+3​,u2​=2⋅2−(−5)−3​
u=2⋅2−(−5)+3​:2
2⋅2−(−5)+3​
Apply rule −(−a)=a=2⋅25+3​
Add the numbers: 5+3=8=2⋅28​
Multiply the numbers: 2⋅2=4=48​
Divide the numbers: 48​=2=2
u=2⋅2−(−5)−3​:21​
2⋅2−(−5)−3​
Apply rule −(−a)=a=2⋅25−3​
Subtract the numbers: 5−3=2=2⋅22​
Multiply the numbers: 2⋅2=4=42​
Cancel the common factor: 2=21​
The solutions to the quadratic equation are:u=2,u=21​
Substitute back u=csc(3x)csc(3x)=2,csc(3x)=21​
csc(3x)=2,csc(3x)=21​
csc(3x)=2:x=18π​+32πn​,x=185π​+32πn​
csc(3x)=2
General solutions for csc(3x)=2
csc(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​csc(x)Undefiend22​323​​1323​​2​2​xπ67π​45π​34π​23π​35π​47π​611π​​csc(x)Undefiend−2−2​−323​​−1−323​​−2​−2​​
3x=6π​+2πn,3x=65π​+2πn
3x=6π​+2πn,3x=65π​+2πn
Solve 3x=6π​+2πn:x=18π​+32πn​
3x=6π​+2πn
Divide both sides by 3
3x=6π​+2πn
Divide both sides by 333x​=36π​​+32πn​
Simplify
33x​=36π​​+32πn​
Simplify 33x​:x
33x​
Divide the numbers: 33​=1=x
Simplify 36π​​+32πn​:18π​+32πn​
36π​​+32πn​
36π​​=18π​
36π​​
Apply the fraction rule: acb​​=c⋅ab​=6⋅3π​
Multiply the numbers: 6⋅3=18=18π​
=18π​+32πn​
x=18π​+32πn​
x=18π​+32πn​
x=18π​+32πn​
Solve 3x=65π​+2πn:x=185π​+32πn​
3x=65π​+2πn
Divide both sides by 3
3x=65π​+2πn
Divide both sides by 333x​=365π​​+32πn​
Simplify
33x​=365π​​+32πn​
Simplify 33x​:x
33x​
Divide the numbers: 33​=1=x
Simplify 365π​​+32πn​:185π​+32πn​
365π​​+32πn​
365π​​=185π​
365π​​
Apply the fraction rule: acb​​=c⋅ab​=6⋅35π​
Multiply the numbers: 6⋅3=18=185π​
=185π​+32πn​
x=185π​+32πn​
x=185π​+32πn​
x=185π​+32πn​
x=18π​+32πn​,x=185π​+32πn​
csc(3x)=21​:No Solution
csc(3x)=21​
csc(x)≤−1orcsc(x)≥1NoSolution
Combine all the solutionsx=18π​+32πn​,x=185π​+32πn​

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Frequently Asked Questions (FAQ)

  • What is the general solution for 2cot^2(3x)=5csc(3x)-4 ?

    The general solution for 2cot^2(3x)=5csc(3x)-4 is x= pi/(18)+(2pin)/3 ,x=(5pi}{18}+\frac{2pin)/3
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