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Popular Trigonometry >

4sin(x)+1=3csc(x)

  • Pre Algebra
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Solution

4sin(x)+1=3csc(x)

Solution

x=23π​+2πn,x=0.84806…+2πn,x=π−0.84806…+2πn
+1
Degrees
x=270∘+360∘n,x=48.59037…∘+360∘n,x=131.40962…∘+360∘n
Solution steps
4sin(x)+1=3csc(x)
Subtract 3csc(x) from both sides4sin(x)+1−3csc(x)=0
Rewrite using trig identities
1−3csc(x)+4sin(x)
Use the basic trigonometric identity: sin(x)=csc(x)1​=1−3csc(x)+4⋅csc(x)1​
4⋅csc(x)1​=csc(x)4​
4⋅csc(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=csc(x)1⋅4​
Multiply the numbers: 1⋅4=4=csc(x)4​
=1−3csc(x)+csc(x)4​
1+csc(x)4​−3csc(x)=0
Solve by substitution
1+csc(x)4​−3csc(x)=0
Let: csc(x)=u1+u4​−3u=0
1+u4​−3u=0:u=−1,u=34​
1+u4​−3u=0
Multiply both sides by u
1+u4​−3u=0
Multiply both sides by u1⋅u+u4​u−3uu=0⋅u
Simplify
1⋅u+u4​u−3uu=0⋅u
Simplify 1⋅u:u
1⋅u
Multiply: 1⋅u=u=u
Simplify u4​u:4
u4​u
Multiply fractions: a⋅cb​=ca⋅b​=u4u​
Cancel the common factor: u=4
Simplify −3uu:−3u2
−3uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=−3u1+1
Add the numbers: 1+1=2=−3u2
Simplify 0⋅u:0
0⋅u
Apply rule 0⋅a=0=0
u+4−3u2=0
u+4−3u2=0
u+4−3u2=0
Solve u+4−3u2=0:u=−1,u=34​
u+4−3u2=0
Write in the standard form ax2+bx+c=0−3u2+u+4=0
Solve with the quadratic formula
−3u2+u+4=0
Quadratic Equation Formula:
For a=−3,b=1,c=4u1,2​=2(−3)−1±12−4(−3)⋅4​​
u1,2​=2(−3)−1±12−4(−3)⋅4​​
12−4(−3)⋅4​=7
12−4(−3)⋅4​
Apply rule 1a=112=1=1−4(−3)⋅4​
Apply rule −(−a)=a=1+4⋅3⋅4​
Multiply the numbers: 4⋅3⋅4=48=1+48​
Add the numbers: 1+48=49=49​
Factor the number: 49=72=72​
Apply radical rule: 72​=7=7
u1,2​=2(−3)−1±7​
Separate the solutionsu1​=2(−3)−1+7​,u2​=2(−3)−1−7​
u=2(−3)−1+7​:−1
2(−3)−1+7​
Remove parentheses: (−a)=−a=−2⋅3−1+7​
Add/Subtract the numbers: −1+7=6=−2⋅36​
Multiply the numbers: 2⋅3=6=−66​
Apply the fraction rule: −ba​=−ba​=−66​
Apply rule aa​=1=−1
u=2(−3)−1−7​:34​
2(−3)−1−7​
Remove parentheses: (−a)=−a=−2⋅3−1−7​
Subtract the numbers: −1−7=−8=−2⋅3−8​
Multiply the numbers: 2⋅3=6=−6−8​
Apply the fraction rule: −b−a​=ba​=68​
Cancel the common factor: 2=34​
The solutions to the quadratic equation are:u=−1,u=34​
u=−1,u=34​
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of 1+u4​−3u and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=−1,u=34​
Substitute back u=csc(x)csc(x)=−1,csc(x)=34​
csc(x)=−1,csc(x)=34​
csc(x)=−1:x=23π​+2πn
csc(x)=−1
General solutions for csc(x)=−1
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​​
x=23π​+2πn
x=23π​+2πn
csc(x)=34​:x=arccsc(34​)+2πn,x=π−arccsc(34​)+2πn
csc(x)=34​
Apply trig inverse properties
csc(x)=34​
General solutions for csc(x)=34​csc(x)=a⇒x=arccsc(a)+2πn,x=π−arccsc(a)+2πnx=arccsc(34​)+2πn,x=π−arccsc(34​)+2πn
x=arccsc(34​)+2πn,x=π−arccsc(34​)+2πn
Combine all the solutionsx=23π​+2πn,x=arccsc(34​)+2πn,x=π−arccsc(34​)+2πn
Show solutions in decimal formx=23π​+2πn,x=0.84806…+2πn,x=π−0.84806…+2πn

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Popular Examples

cos^2(x)-3sin(x)=0csc(t)= 1/(-(sqrt(2))/2)cos^2(x)=1-sin(x),0<= x<2picos(2x)-3sin(x)=2,0<= x<= 3602cos^2(x)+3cos(x)=4cos(x)+1

Frequently Asked Questions (FAQ)

  • What is the general solution for 4sin(x)+1=3csc(x) ?

    The general solution for 4sin(x)+1=3csc(x) is x=(3pi)/2+2pin,x=0.84806…+2pin,x=pi-0.84806…+2pin
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