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Popular Trigonometry >

4sin(2x)+csc(2x)=4

  • Pre Algebra
  • Algebra
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Solution

4sin(2x)+csc(2x)=4

Solution

x=12π​+πn,x=125π​+πn
+1
Degrees
x=15∘+180∘n,x=75∘+180∘n
Solution steps
4sin(2x)+csc(2x)=4
Subtract 4 from both sides4sin(2x)+csc(2x)−4=0
Rewrite using trig identities
−4+csc(2x)+4sin(2x)
Use the basic trigonometric identity: sin(x)=csc(x)1​=−4+csc(2x)+4⋅csc(2x)1​
4⋅csc(2x)1​=csc(2x)4​
4⋅csc(2x)1​
Multiply fractions: a⋅cb​=ca⋅b​=csc(2x)1⋅4​
Multiply the numbers: 1⋅4=4=csc(2x)4​
=−4+csc(2x)+csc(2x)4​
−4+csc(2x)+csc(2x)4​=0
Solve by substitution
−4+csc(2x)+csc(2x)4​=0
Let: csc(2x)=u−4+u+u4​=0
−4+u+u4​=0:u=2
−4+u+u4​=0
Multiply both sides by u
−4+u+u4​=0
Multiply both sides by u−4u+uu+u4​u=0⋅u
Simplify
−4u+uu+u4​u=0⋅u
Simplify uu:u2
uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=u1+1
Add the numbers: 1+1=2=u2
Simplify u4​u:4
u4​u
Multiply fractions: a⋅cb​=ca⋅b​=u4u​
Cancel the common factor: u=4
Simplify 0⋅u:0
0⋅u
Apply rule 0⋅a=0=0
−4u+u2+4=0
−4u+u2+4=0
−4u+u2+4=0
Solve −4u+u2+4=0:u=2
−4u+u2+4=0
Write in the standard form ax2+bx+c=0u2−4u+4=0
Solve with the quadratic formula
u2−4u+4=0
Quadratic Equation Formula:
For a=1,b=−4,c=4u1,2​=2⋅1−(−4)±(−4)2−4⋅1⋅4​​
u1,2​=2⋅1−(−4)±(−4)2−4⋅1⋅4​​
(−4)2−4⋅1⋅4=0
(−4)2−4⋅1⋅4
Apply exponent rule: (−a)n=an,if n is even(−4)2=42=42−4⋅1⋅4
Multiply the numbers: 4⋅1⋅4=16=42−16
42=16=16−16
Subtract the numbers: 16−16=0=0
u1,2​=2⋅1−(−4)±0​​
u=2⋅1−(−4)​
2⋅1−(−4)​=2
2⋅1−(−4)​
Apply rule −(−a)=a=2⋅14​
Multiply the numbers: 2⋅1=2=24​
Divide the numbers: 24​=2=2
u=2
The solution to the quadratic equation is:u=2
u=2
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −4+u+u4​ and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=2
Substitute back u=csc(2x)csc(2x)=2
csc(2x)=2
csc(2x)=2:x=12π​+πn,x=125π​+πn
csc(2x)=2
General solutions for csc(2x)=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​​
2x=6π​+2πn,2x=65π​+2πn
2x=6π​+2πn,2x=65π​+2πn
Solve 2x=6π​+2πn:x=12π​+πn
2x=6π​+2πn
Divide both sides by 2
2x=6π​+2πn
Divide both sides by 222x​=26π​​+22πn​
Simplify
22x​=26π​​+22πn​
Simplify 22x​:x
22x​
Divide the numbers: 22​=1=x
Simplify 26π​​+22πn​:12π​+πn
26π​​+22πn​
26π​​=12π​
26π​​
Apply the fraction rule: acb​​=c⋅ab​=6⋅2π​
Multiply the numbers: 6⋅2=12=12π​
22πn​=πn
22πn​
Divide the numbers: 22​=1=πn
=12π​+πn
x=12π​+πn
x=12π​+πn
x=12π​+πn
Solve 2x=65π​+2πn:x=125π​+πn
2x=65π​+2πn
Divide both sides by 2
2x=65π​+2πn
Divide both sides by 222x​=265π​​+22πn​
Simplify
22x​=265π​​+22πn​
Simplify 22x​:x
22x​
Divide the numbers: 22​=1=x
Simplify 265π​​+22πn​:125π​+πn
265π​​+22πn​
265π​​=125π​
265π​​
Apply the fraction rule: acb​​=c⋅ab​=6⋅25π​
Multiply the numbers: 6⋅2=12=125π​
22πn​=πn
22πn​
Divide the numbers: 22​=1=πn
=125π​+πn
x=125π​+πn
x=125π​+πn
x=125π​+πn
x=12π​+πn,x=125π​+πn
Combine all the solutionsx=12π​+πn,x=125π​+πn

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Popular Examples

cot(θ)= 9/40sqrt(5)cos(x)-2sin(x)=0.5sin(x)= 6/15cot(θ)=1.475/125 =(cosh(0.2x))/(cosh(0.4x))

Frequently Asked Questions (FAQ)

  • What is the general solution for 4sin(2x)+csc(2x)=4 ?

    The general solution for 4sin(2x)+csc(2x)=4 is x= pi/(12)+pin,x=(5pi)/(12)+pin
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