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Popular Trigonometry >

(1+4cos(θ))^2=(sqrt(3)sin(θ))^2

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Solution

(1+4cos(θ))2=(3​sin(θ))2

Solution

θ=1.39363…+2πn,θ=2π−1.39363…+2πn,θ=2.21091…+2πn,θ=−2.21091…+2πn
+1
Degrees
θ=79.84945…∘+360∘n,θ=280.15054…∘+360∘n,θ=126.67590…∘+360∘n,θ=−126.67590…∘+360∘n
Solution steps
(1+4cos(θ))2=(3​sin(θ))2
Subtract (3​sin(θ))2 from both sides(1+4cos(θ))2−3sin2(θ)=0
Rewrite using trig identities
(1+4cos(θ))2−3sin2(θ)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(1+4cos(θ))2−3(1−cos2(θ))
Simplify (1+4cos(θ))2−3(1−cos2(θ)):19cos2(θ)+8cos(θ)−2
(1+4cos(θ))2−3(1−cos2(θ))
(1+4cos(θ))2:1+8cos(θ)+16cos2(θ)
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=1,b=4cos(θ)
=12+2⋅1⋅4cos(θ)+(4cos(θ))2
Simplify 12+2⋅1⋅4cos(θ)+(4cos(θ))2:1+8cos(θ)+16cos2(θ)
12+2⋅1⋅4cos(θ)+(4cos(θ))2
Apply rule 1a=112=1=1+2⋅1⋅4cos(θ)+(4cos(θ))2
2⋅1⋅4cos(θ)=8cos(θ)
2⋅1⋅4cos(θ)
Multiply the numbers: 2⋅1⋅4=8=8cos(θ)
(4cos(θ))2=16cos2(θ)
(4cos(θ))2
Apply exponent rule: (a⋅b)n=anbn=42cos2(θ)
42=16=16cos2(θ)
=1+8cos(θ)+16cos2(θ)
=1+8cos(θ)+16cos2(θ)
=1+8cos(θ)+16cos2(θ)−3(1−cos2(θ))
Expand −3(1−cos2(θ)):−3+3cos2(θ)
−3(1−cos2(θ))
Apply the distributive law: a(b−c)=ab−aca=−3,b=1,c=cos2(θ)=−3⋅1−(−3)cos2(θ)
Apply minus-plus rules−(−a)=a=−3⋅1+3cos2(θ)
Multiply the numbers: 3⋅1=3=−3+3cos2(θ)
=1+8cos(θ)+16cos2(θ)−3+3cos2(θ)
Simplify 1+8cos(θ)+16cos2(θ)−3+3cos2(θ):19cos2(θ)+8cos(θ)−2
1+8cos(θ)+16cos2(θ)−3+3cos2(θ)
Group like terms=8cos(θ)+16cos2(θ)+3cos2(θ)+1−3
Add similar elements: 16cos2(θ)+3cos2(θ)=19cos2(θ)=8cos(θ)+19cos2(θ)+1−3
Add/Subtract the numbers: 1−3=−2=19cos2(θ)+8cos(θ)−2
=19cos2(θ)+8cos(θ)−2
=19cos2(θ)+8cos(θ)−2
−2+19cos2(θ)+8cos(θ)=0
Solve by substitution
−2+19cos2(θ)+8cos(θ)=0
Let: cos(θ)=u−2+19u2+8u=0
−2+19u2+8u=0:u=19−4+36​​,u=−194+36​​
−2+19u2+8u=0
Write in the standard form ax2+bx+c=019u2+8u−2=0
Solve with the quadratic formula
19u2+8u−2=0
Quadratic Equation Formula:
For a=19,b=8,c=−2u1,2​=2⋅19−8±82−4⋅19(−2)​​
u1,2​=2⋅19−8±82−4⋅19(−2)​​
82−4⋅19(−2)​=66​
82−4⋅19(−2)​
Apply rule −(−a)=a=82+4⋅19⋅2​
Multiply the numbers: 4⋅19⋅2=152=82+152​
82=64=64+152​
Add the numbers: 64+152=216=216​
Prime factorization of 216:23⋅33
216
216divides by 2216=108⋅2=2⋅108
108divides by 2108=54⋅2=2⋅2⋅54
54divides by 254=27⋅2=2⋅2⋅2⋅27
27divides by 327=9⋅3=2⋅2⋅2⋅3⋅9
9divides by 39=3⋅3=2⋅2⋅2⋅3⋅3⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅2⋅3⋅3⋅3
=23⋅33
=23⋅33​
Apply exponent rule: ab+c=ab⋅ac=22⋅32⋅2⋅3​
Apply radical rule: =22​32​2⋅3​
Apply radical rule: 22​=2=232​2⋅3​
Apply radical rule: 32​=3=2⋅32⋅3​
Refine=66​
u1,2​=2⋅19−8±66​​
Separate the solutionsu1​=2⋅19−8+66​​,u2​=2⋅19−8−66​​
u=2⋅19−8+66​​:19−4+36​​
2⋅19−8+66​​
Multiply the numbers: 2⋅19=38=38−8+66​​
Factor −8+66​:2(−4+36​)
−8+66​
Rewrite as=−2⋅4+2⋅36​
Factor out common term 2=2(−4+36​)
=382(−4+36​)​
Cancel the common factor: 2=19−4+36​​
u=2⋅19−8−66​​:−194+36​​
2⋅19−8−66​​
Multiply the numbers: 2⋅19=38=38−8−66​​
Factor −8−66​:−2(4+36​)
−8−66​
Rewrite as=−2⋅4−2⋅36​
Factor out common term 2=−2(4+36​)
=−382(4+36​)​
Cancel the common factor: 2=−194+36​​
The solutions to the quadratic equation are:u=19−4+36​​,u=−194+36​​
Substitute back u=cos(θ)cos(θ)=19−4+36​​,cos(θ)=−194+36​​
cos(θ)=19−4+36​​,cos(θ)=−194+36​​
cos(θ)=19−4+36​​:θ=arccos(19−4+36​​)+2πn,θ=2π−arccos(19−4+36​​)+2πn
cos(θ)=19−4+36​​
Apply trig inverse properties
cos(θ)=19−4+36​​
General solutions for cos(θ)=19−4+36​​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(19−4+36​​)+2πn,θ=2π−arccos(19−4+36​​)+2πn
θ=arccos(19−4+36​​)+2πn,θ=2π−arccos(19−4+36​​)+2πn
cos(θ)=−194+36​​:θ=arccos(−194+36​​)+2πn,θ=−arccos(−194+36​​)+2πn
cos(θ)=−194+36​​
Apply trig inverse properties
cos(θ)=−194+36​​
General solutions for cos(θ)=−194+36​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−194+36​​)+2πn,θ=−arccos(−194+36​​)+2πn
θ=arccos(−194+36​​)+2πn,θ=−arccos(−194+36​​)+2πn
Combine all the solutionsθ=arccos(19−4+36​​)+2πn,θ=2π−arccos(19−4+36​​)+2πn,θ=arccos(−194+36​​)+2πn,θ=−arccos(−194+36​​)+2πn
Show solutions in decimal formθ=1.39363…+2πn,θ=2π−1.39363…+2πn,θ=2.21091…+2πn,θ=−2.21091…+2πn

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