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Popular Trigonometry >

15cos^2(θ)+2sin^2(θ)=7

  • Pre Algebra
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Solution

15cos2(θ)+2sin2(θ)=7

Solution

θ=0.90183…+2πn,θ=π−0.90183…+2πn,θ=−0.90183…+2πn,θ=π+0.90183…+2πn
+1
Degrees
θ=51.67118…∘+360∘n,θ=128.32881…∘+360∘n,θ=−51.67118…∘+360∘n,θ=231.67118…∘+360∘n
Solution steps
15cos2(θ)+2sin2(θ)=7
Subtract 7 from both sides15cos2(θ)+2sin2(θ)−7=0
Rewrite using trig identities
−7+15cos2(θ)+2sin2(θ)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−7+15(1−sin2(θ))+2sin2(θ)
Simplify −7+15(1−sin2(θ))+2sin2(θ):−13sin2(θ)+8
−7+15(1−sin2(θ))+2sin2(θ)
Expand 15(1−sin2(θ)):15−15sin2(θ)
15(1−sin2(θ))
Apply the distributive law: a(b−c)=ab−aca=15,b=1,c=sin2(θ)=15⋅1−15sin2(θ)
Multiply the numbers: 15⋅1=15=15−15sin2(θ)
=−7+15−15sin2(θ)+2sin2(θ)
Simplify −7+15−15sin2(θ)+2sin2(θ):−13sin2(θ)+8
−7+15−15sin2(θ)+2sin2(θ)
Add similar elements: −15sin2(θ)+2sin2(θ)=−13sin2(θ)=−7+15−13sin2(θ)
Add/Subtract the numbers: −7+15=8=−13sin2(θ)+8
=−13sin2(θ)+8
=−13sin2(θ)+8
8−13sin2(θ)=0
Solve by substitution
8−13sin2(θ)=0
Let: sin(θ)=u8−13u2=0
8−13u2=0:u=13226​​,u=−13226​​
8−13u2=0
Move 8to the right side
8−13u2=0
Subtract 8 from both sides8−13u2−8=0−8
Simplify−13u2=−8
−13u2=−8
Divide both sides by −13
−13u2=−8
Divide both sides by −13−13−13u2​=−13−8​
Simplifyu2=138​
u2=138​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=138​​,u=−138​​
138​​=13226​​
138​​
Apply radical rule: assuming a≥0,b≥0=13​8​​
8​=22​
8​
Prime factorization of 8:23
8
8divides by 28=4⋅2=2⋅4
4divides by 24=2⋅2=2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2
=23
=23​
Apply exponent rule: ab+c=ab⋅ac=22⋅2​
Apply radical rule: =2​22​
Apply radical rule: 22​=2=22​
=13​22​​
Rationalize 13​22​​:13226​​
13​22​​
Multiply by the conjugate 13​13​​=13​13​22​13​​
22​13​=226​
22​13​
Apply radical rule: a​b​=a⋅b​2​13​=2⋅13​=22⋅13​
Multiply the numbers: 2⋅13=26=226​
13​13​=13
13​13​
Apply radical rule: a​a​=a13​13​=13=13
=13226​​
=13226​​
−138​​=−13226​​
−138​​
Simplify 138​​:13​22​​
138​​
Apply radical rule: assuming a≥0,b≥0=13​8​​
8​=22​
8​
Prime factorization of 8:23
8
8divides by 28=4⋅2=2⋅4
4divides by 24=2⋅2=2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2
=23
=23​
Apply exponent rule: ab+c=ab⋅ac=22⋅2​
Apply radical rule: =2​22​
Apply radical rule: 22​=2=22​
=13​22​​
=−13​22​​
Rationalize −13​22​​:−13226​​
−13​22​​
Multiply by the conjugate 13​13​​=−13​13​22​13​​
22​13​=226​
22​13​
Apply radical rule: a​b​=a⋅b​2​13​=2⋅13​=22⋅13​
Multiply the numbers: 2⋅13=26=226​
13​13​=13
13​13​
Apply radical rule: a​a​=a13​13​=13=13
=−13226​​
=−13226​​
u=13226​​,u=−13226​​
Substitute back u=sin(θ)sin(θ)=13226​​,sin(θ)=−13226​​
sin(θ)=13226​​,sin(θ)=−13226​​
sin(θ)=13226​​:θ=arcsin(13226​​)+2πn,θ=π−arcsin(13226​​)+2πn
sin(θ)=13226​​
Apply trig inverse properties
sin(θ)=13226​​
General solutions for sin(θ)=13226​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(13226​​)+2πn,θ=π−arcsin(13226​​)+2πn
θ=arcsin(13226​​)+2πn,θ=π−arcsin(13226​​)+2πn
sin(θ)=−13226​​:θ=arcsin(−13226​​)+2πn,θ=π+arcsin(13226​​)+2πn
sin(θ)=−13226​​
Apply trig inverse properties
sin(θ)=−13226​​
General solutions for sin(θ)=−13226​​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnθ=arcsin(−13226​​)+2πn,θ=π+arcsin(13226​​)+2πn
θ=arcsin(−13226​​)+2πn,θ=π+arcsin(13226​​)+2πn
Combine all the solutionsθ=arcsin(13226​​)+2πn,θ=π−arcsin(13226​​)+2πn,θ=arcsin(−13226​​)+2πn,θ=π+arcsin(13226​​)+2πn
Show solutions in decimal formθ=0.90183…+2πn,θ=π−0.90183…+2πn,θ=−0.90183…+2πn,θ=π+0.90183…+2πn

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