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Popular Trigonometry >

3sin(θ)-cos(2θ)=1

  • Pre Algebra
  • Algebra
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Solution

3sin(θ)−cos(2θ)=1

Solution

θ=6π​+2πn,θ=65π​+2πn
+1
Degrees
θ=30∘+360∘n,θ=150∘+360∘n
Solution steps
3sin(θ)−cos(2θ)=1
Subtract 1 from both sides3sin(θ)−cos(2θ)−1=0
Rewrite using trig identities
−1−cos(2θ)+3sin(θ)
Use the Double Angle identity: cos(2x)=1−2sin2(x)=−1−(1−2sin2(θ))+3sin(θ)
Simplify −1−(1−2sin2(θ))+3sin(θ):2sin2(θ)+3sin(θ)−2
−1−(1−2sin2(θ))+3sin(θ)
−(1−2sin2(θ)):−1+2sin2(θ)
−(1−2sin2(θ))
Distribute parentheses=−(1)−(−2sin2(θ))
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+2sin2(θ)
=−1−1+2sin2(θ)+3sin(θ)
Subtract the numbers: −1−1=−2=2sin2(θ)+3sin(θ)−2
=2sin2(θ)+3sin(θ)−2
−2+2sin2(θ)+3sin(θ)=0
Solve by substitution
−2+2sin2(θ)+3sin(θ)=0
Let: sin(θ)=u−2+2u2+3u=0
−2+2u2+3u=0:u=21​,u=−2
−2+2u2+3u=0
Write in the standard form ax2+bx+c=02u2+3u−2=0
Solve with the quadratic formula
2u2+3u−2=0
Quadratic Equation Formula:
For a=2,b=3,c=−2u1,2​=2⋅2−3±32−4⋅2(−2)​​
u1,2​=2⋅2−3±32−4⋅2(−2)​​
32−4⋅2(−2)​=5
32−4⋅2(−2)​
Apply rule −(−a)=a=32+4⋅2⋅2​
Multiply the numbers: 4⋅2⋅2=16=32+16​
32=9=9+16​
Add the numbers: 9+16=25=25​
Factor the number: 25=52=52​
Apply radical rule: 52​=5=5
u1,2​=2⋅2−3±5​
Separate the solutionsu1​=2⋅2−3+5​,u2​=2⋅2−3−5​
u=2⋅2−3+5​:21​
2⋅2−3+5​
Add/Subtract the numbers: −3+5=2=2⋅22​
Multiply the numbers: 2⋅2=4=42​
Cancel the common factor: 2=21​
u=2⋅2−3−5​:−2
2⋅2−3−5​
Subtract the numbers: −3−5=−8=2⋅2−8​
Multiply the numbers: 2⋅2=4=4−8​
Apply the fraction rule: b−a​=−ba​=−48​
Divide the numbers: 48​=2=−2
The solutions to the quadratic equation are:u=21​,u=−2
Substitute back u=sin(θ)sin(θ)=21​,sin(θ)=−2
sin(θ)=21​,sin(θ)=−2
sin(θ)=21​:θ=6π​+2πn,θ=65π​+2πn
sin(θ)=21​
General solutions for sin(θ)=21​
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
θ=6π​+2πn,θ=65π​+2πn
θ=6π​+2πn,θ=65π​+2πn
sin(θ)=−2:No Solution
sin(θ)=−2
−1≤sin(x)≤1NoSolution
Combine all the solutionsθ=6π​+2πn,θ=65π​+2πn

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Popular Examples

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

Frequently Asked Questions (FAQ)

  • What is the general solution for 3sin(θ)-cos(2θ)=1 ?

    The general solution for 3sin(θ)-cos(2θ)=1 is θ= pi/6+2pin,θ=(5pi)/6+2pin
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