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Popular Trigonometry >

2-3sin(θ)=cos(θ)

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Solution

2−3sin(θ)=cos(θ)

Solution

θ=π−1.00646…+2πn,θ=0.36296…+2πn
+1
Degrees
θ=122.33353…∘+360∘n,θ=20.79657…∘+360∘n
Solution steps
2−3sin(θ)=cos(θ)
Square both sides(2−3sin(θ))2=cos2(θ)
Subtract cos2(θ) from both sides(2−3sin(θ))2−cos2(θ)=0
Rewrite using trig identities
(2−3sin(θ))2−cos2(θ)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=(2−3sin(θ))2−(1−sin2(θ))
Simplify (2−3sin(θ))2−(1−sin2(θ)):10sin2(θ)−12sin(θ)+3
(2−3sin(θ))2−(1−sin2(θ))
(2−3sin(θ))2:4−12sin(θ)+9sin2(θ)
Apply Perfect Square Formula: (a−b)2=a2−2ab+b2a=2,b=3sin(θ)
=22−2⋅2⋅3sin(θ)+(3sin(θ))2
Simplify 22−2⋅2⋅3sin(θ)+(3sin(θ))2:4−12sin(θ)+9sin2(θ)
22−2⋅2⋅3sin(θ)+(3sin(θ))2
22=4
22
22=4=4
2⋅2⋅3sin(θ)=12sin(θ)
2⋅2⋅3sin(θ)
Multiply the numbers: 2⋅2⋅3=12=12sin(θ)
(3sin(θ))2=9sin2(θ)
(3sin(θ))2
Apply exponent rule: (a⋅b)n=anbn=32sin2(θ)
32=9=9sin2(θ)
=4−12sin(θ)+9sin2(θ)
=4−12sin(θ)+9sin2(θ)
=4−12sin(θ)+9sin2(θ)−(1−sin2(θ))
−(1−sin2(θ)):−1+sin2(θ)
−(1−sin2(θ))
Distribute parentheses=−(1)−(−sin2(θ))
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+sin2(θ)
=4−12sin(θ)+9sin2(θ)−1+sin2(θ)
Simplify 4−12sin(θ)+9sin2(θ)−1+sin2(θ):10sin2(θ)−12sin(θ)+3
4−12sin(θ)+9sin2(θ)−1+sin2(θ)
Group like terms=−12sin(θ)+9sin2(θ)+sin2(θ)+4−1
Add similar elements: 9sin2(θ)+sin2(θ)=10sin2(θ)=−12sin(θ)+10sin2(θ)+4−1
Add/Subtract the numbers: 4−1=3=10sin2(θ)−12sin(θ)+3
=10sin2(θ)−12sin(θ)+3
=10sin2(θ)−12sin(θ)+3
3+10sin2(θ)−12sin(θ)=0
Solve by substitution
3+10sin2(θ)−12sin(θ)=0
Let: sin(θ)=u3+10u2−12u=0
3+10u2−12u=0:u=106+6​​,u=106−6​​
3+10u2−12u=0
Write in the standard form ax2+bx+c=010u2−12u+3=0
Solve with the quadratic formula
10u2−12u+3=0
Quadratic Equation Formula:
For a=10,b=−12,c=3u1,2​=2⋅10−(−12)±(−12)2−4⋅10⋅3​​
u1,2​=2⋅10−(−12)±(−12)2−4⋅10⋅3​​
(−12)2−4⋅10⋅3​=26​
(−12)2−4⋅10⋅3​
Apply exponent rule: (−a)n=an,if n is even(−12)2=122=122−4⋅10⋅3​
Multiply the numbers: 4⋅10⋅3=120=122−120​
122=144=144−120​
Subtract the numbers: 144−120=24=24​
Prime factorization of 24:23⋅3
24
24divides by 224=12⋅2=2⋅12
12divides by 212=6⋅2=2⋅2⋅6
6divides by 26=3⋅2=2⋅2⋅2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅2⋅3
=23⋅3
=23⋅3​
Apply exponent rule: ab+c=ab⋅ac=22⋅2⋅3​
Apply radical rule: =22​2⋅3​
Apply radical rule: 22​=2=22⋅3​
Refine=26​
u1,2​=2⋅10−(−12)±26​​
Separate the solutionsu1​=2⋅10−(−12)+26​​,u2​=2⋅10−(−12)−26​​
u=2⋅10−(−12)+26​​:106+6​​
2⋅10−(−12)+26​​
Apply rule −(−a)=a=2⋅1012+26​​
Multiply the numbers: 2⋅10=20=2012+26​​
Factor 12+26​:2(6+6​)
12+26​
Rewrite as=2⋅6+26​
Factor out common term 2=2(6+6​)
=202(6+6​)​
Cancel the common factor: 2=106+6​​
u=2⋅10−(−12)−26​​:106−6​​
2⋅10−(−12)−26​​
Apply rule −(−a)=a=2⋅1012−26​​
Multiply the numbers: 2⋅10=20=2012−26​​
Factor 12−26​:2(6−6​)
12−26​
Rewrite as=2⋅6−26​
Factor out common term 2=2(6−6​)
=202(6−6​)​
Cancel the common factor: 2=106−6​​
The solutions to the quadratic equation are:u=106+6​​,u=106−6​​
Substitute back u=sin(θ)sin(θ)=106+6​​,sin(θ)=106−6​​
sin(θ)=106+6​​,sin(θ)=106−6​​
sin(θ)=106+6​​:θ=arcsin(106+6​​)+2πn,θ=π−arcsin(106+6​​)+2πn
sin(θ)=106+6​​
Apply trig inverse properties
sin(θ)=106+6​​
General solutions for sin(θ)=106+6​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(106+6​​)+2πn,θ=π−arcsin(106+6​​)+2πn
θ=arcsin(106+6​​)+2πn,θ=π−arcsin(106+6​​)+2πn
sin(θ)=106−6​​:θ=arcsin(106−6​​)+2πn,θ=π−arcsin(106−6​​)+2πn
sin(θ)=106−6​​
Apply trig inverse properties
sin(θ)=106−6​​
General solutions for sin(θ)=106−6​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(106−6​​)+2πn,θ=π−arcsin(106−6​​)+2πn
θ=arcsin(106−6​​)+2πn,θ=π−arcsin(106−6​​)+2πn
Combine all the solutionsθ=arcsin(106+6​​)+2πn,θ=π−arcsin(106+6​​)+2πn,θ=arcsin(106−6​​)+2πn,θ=π−arcsin(106−6​​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into 2−3sin(θ)=cos(θ)
Remove the ones that don't agree with the equation.
Check the solution arcsin(106+6​​)+2πn:False
arcsin(106+6​​)+2πn
Plug in n=1arcsin(106+6​​)+2π1
For 2−3sin(θ)=cos(θ)plug inθ=arcsin(106+6​​)+2π12−3sin(arcsin(106+6​​)+2π1)=cos(arcsin(106+6​​)+2π1)
Refine−0.53484…=0.53484…
⇒False
Check the solution π−arcsin(106+6​​)+2πn:True
π−arcsin(106+6​​)+2πn
Plug in n=1π−arcsin(106+6​​)+2π1
For 2−3sin(θ)=cos(θ)plug inθ=π−arcsin(106+6​​)+2π12−3sin(π−arcsin(106+6​​)+2π1)=cos(π−arcsin(106+6​​)+2π1)
Refine−0.53484…=−0.53484…
⇒True
Check the solution arcsin(106−6​​)+2πn:True
arcsin(106−6​​)+2πn
Plug in n=1arcsin(106−6​​)+2π1
For 2−3sin(θ)=cos(θ)plug inθ=arcsin(106−6​​)+2π12−3sin(arcsin(106−6​​)+2π1)=cos(arcsin(106−6​​)+2π1)
Refine0.93484…=0.93484…
⇒True
Check the solution π−arcsin(106−6​​)+2πn:False
π−arcsin(106−6​​)+2πn
Plug in n=1π−arcsin(106−6​​)+2π1
For 2−3sin(θ)=cos(θ)plug inθ=π−arcsin(106−6​​)+2π12−3sin(π−arcsin(106−6​​)+2π1)=cos(π−arcsin(106−6​​)+2π1)
Refine0.93484…=−0.93484…
⇒False
θ=π−arcsin(106+6​​)+2πn,θ=arcsin(106−6​​)+2πn
Show solutions in decimal formθ=π−1.00646…+2πn,θ=0.36296…+2πn

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Frequently Asked Questions (FAQ)

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

    The general solution for 2-3sin(θ)=cos(θ) is θ=pi-1.00646…+2pin,θ=0.36296…+2pin
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