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Popular Trigonometry >

5cos^2(θ)+3sqrt(2)sin(θ)-11/2 =0

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Solution

5cos2(θ)+32​sin(θ)−211​=0

Solution

θ=0.14189…+2πn,θ=π−0.14189…+2πn,θ=4π​+2πn,θ=43π​+2πn
+1
Degrees
θ=8.13010…∘+360∘n,θ=171.86989…∘+360∘n,θ=45∘+360∘n,θ=135∘+360∘n
Solution steps
5cos2(θ)+32​sin(θ)−211​=0
Simplify 5cos2(θ)+32​sin(θ)−211​:210cos2(θ)+62​sin(θ)−11​
5cos2(θ)+32​sin(θ)−211​
Convert element to fraction: 5cos2(θ)=25cos2(θ)2​,32​sin(θ)=23⋅2​sin(θ)2​=25cos2(θ)⋅2​+232​sin(θ)⋅2​−211​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=25cos2(θ)⋅2+32​sin(θ)⋅2−11​
5cos2(θ)⋅2+32​sin(θ)⋅2−11=10cos2(θ)+62​sin(θ)−11
5cos2(θ)⋅2+32​sin(θ)⋅2−11
Multiply the numbers: 5⋅2=10=10cos2(θ)+3⋅22​sin(θ)−11
Multiply the numbers: 3⋅2=6=10cos2(θ)+62​sin(θ)−11
=210cos2(θ)+62​sin(θ)−11​
210cos2(θ)+62​sin(θ)−11​=0
g(x)f(x)​=0⇒f(x)=010cos2(θ)+62​sin(θ)−11=0
Rewrite using trig identities
−11+10cos2(θ)+6sin(θ)2​
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−11+10(1−sin2(θ))+6sin(θ)2​
Simplify −11+10(1−sin2(θ))+6sin(θ)2​:62​sin(θ)−10sin2(θ)−1
−11+10(1−sin2(θ))+6sin(θ)2​
=−11+10(1−sin2(θ))+62​sin(θ)
Expand 10(1−sin2(θ)):10−10sin2(θ)
10(1−sin2(θ))
Apply the distributive law: a(b−c)=ab−aca=10,b=1,c=sin2(θ)=10⋅1−10sin2(θ)
Multiply the numbers: 10⋅1=10=10−10sin2(θ)
=−11+10−10sin2(θ)+6sin(θ)2​
Add/Subtract the numbers: −11+10=−1=62​sin(θ)−10sin2(θ)−1
=62​sin(θ)−10sin2(θ)−1
−1−10sin2(θ)+6sin(θ)2​=0
Solve by substitution
−1−10sin2(θ)+6sin(θ)2​=0
Let: sin(θ)=u−1−10u2+6u2​=0
−1−10u2+6u2​=0:u=102​​,u=22​​
−1−10u2+6u2​=0
Write in the standard form ax2+bx+c=0−10u2+62​u−1=0
Solve with the quadratic formula
−10u2+62​u−1=0
Quadratic Equation Formula:
For a=−10,b=62​,c=−1u1,2​=2(−10)−62​±(62​)2−4(−10)(−1)​​
u1,2​=2(−10)−62​±(62​)2−4(−10)(−1)​​
(62​)2−4(−10)(−1)​=42​
(62​)2−4(−10)(−1)​
Apply rule −(−a)=a=(62​)2−4⋅10⋅1​
(62​)2=62⋅2
(62​)2
Apply exponent rule: (a⋅b)n=anbn=62(2​)2
(2​)2:2
Apply radical rule: a​=a21​=(221​)2
Apply exponent rule: (ab)c=abc=221​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=2
=62⋅2
4⋅10⋅1=40
4⋅10⋅1
Multiply the numbers: 4⋅10⋅1=40=40
=62⋅2−40​
62⋅2=72
62⋅2
62=36=36⋅2
Multiply the numbers: 36⋅2=72=72
=72−40​
Subtract the numbers: 72−40=32=32​
Prime factorization of 32:25
32
32divides by 232=16⋅2=2⋅16
16divides by 216=8⋅2=2⋅2⋅8
8divides by 28=4⋅2=2⋅2⋅2⋅4
4divides by 24=2⋅2=2⋅2⋅2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2⋅2⋅2
=25
=25​
Apply exponent rule: ab+c=ab⋅ac=24⋅2​
Apply radical rule: =2​24​
Apply radical rule: 24​=224​=22=222​
Refine=42​
u1,2​=2(−10)−62​±42​​
Separate the solutionsu1​=2(−10)−62​+42​​,u2​=2(−10)−62​−42​​
u=2(−10)−62​+42​​:102​​
2(−10)−62​+42​​
Remove parentheses: (−a)=−a=−2⋅10−62​+42​​
Add similar elements: −62​+42​=−22​=−2⋅10−22​​
Multiply the numbers: 2⋅10=20=−20−22​​
Apply the fraction rule: −b−a​=ba​=2022​​
Cancel the common factor: 2=102​​
u=2(−10)−62​−42​​:22​​
2(−10)−62​−42​​
Remove parentheses: (−a)=−a=−2⋅10−62​−42​​
Add similar elements: −62​−42​=−102​=−2⋅10−102​​
Multiply the numbers: 2⋅10=20=−20−102​​
Apply the fraction rule: −b−a​=ba​=20102​​
Cancel the common factor: 10=22​​
The solutions to the quadratic equation are:u=102​​,u=22​​
Substitute back u=sin(θ)sin(θ)=102​​,sin(θ)=22​​
sin(θ)=102​​,sin(θ)=22​​
sin(θ)=102​​:θ=arcsin(102​​)+2πn,θ=π−arcsin(102​​)+2πn
sin(θ)=102​​
Apply trig inverse properties
sin(θ)=102​​
General solutions for sin(θ)=102​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(102​​)+2πn,θ=π−arcsin(102​​)+2πn
θ=arcsin(102​​)+2πn,θ=π−arcsin(102​​)+2πn
sin(θ)=22​​:θ=4π​+2πn,θ=43π​+2πn
sin(θ)=22​​
General solutions for sin(θ)=22​​
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=4π​+2πn,θ=43π​+2πn
θ=4π​+2πn,θ=43π​+2πn
Combine all the solutionsθ=arcsin(102​​)+2πn,θ=π−arcsin(102​​)+2πn,θ=4π​+2πn,θ=43π​+2πn
Show solutions in decimal formθ=0.14189…+2πn,θ=π−0.14189…+2πn,θ=4π​+2πn,θ=43π​+2πn

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