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Popular Trigonometry >

3cos(x)+sin(x)=1

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Solution

3cos(x)+sin(x)=1

Solution

x=−0.92729…+2πn,x=2π​+2πn
+1
Degrees
x=−53.13010…∘+360∘n,x=90∘+360∘n
Solution steps
3cos(x)+sin(x)=1
Subtract sin(x) from both sides3cos(x)=1−sin(x)
Square both sides(3cos(x))2=(1−sin(x))2
Subtract (1−sin(x))2 from both sides9cos2(x)−1+2sin(x)−sin2(x)=0
Rewrite using trig identities
−1−sin2(x)+2sin(x)+9cos2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−1−sin2(x)+2sin(x)+9(1−sin2(x))
Simplify −1−sin2(x)+2sin(x)+9(1−sin2(x)):2sin(x)−10sin2(x)+8
−1−sin2(x)+2sin(x)+9(1−sin2(x))
Expand 9(1−sin2(x)):9−9sin2(x)
9(1−sin2(x))
Apply the distributive law: a(b−c)=ab−aca=9,b=1,c=sin2(x)=9⋅1−9sin2(x)
Multiply the numbers: 9⋅1=9=9−9sin2(x)
=−1−sin2(x)+2sin(x)+9−9sin2(x)
Simplify −1−sin2(x)+2sin(x)+9−9sin2(x):2sin(x)−10sin2(x)+8
−1−sin2(x)+2sin(x)+9−9sin2(x)
Group like terms=−sin2(x)+2sin(x)−9sin2(x)−1+9
Add similar elements: −sin2(x)−9sin2(x)=−10sin2(x)=−10sin2(x)+2sin(x)−1+9
Add/Subtract the numbers: −1+9=8=2sin(x)−10sin2(x)+8
=2sin(x)−10sin2(x)+8
=2sin(x)−10sin2(x)+8
8−10sin2(x)+2sin(x)=0
Solve by substitution
8−10sin2(x)+2sin(x)=0
Let: sin(x)=u8−10u2+2u=0
8−10u2+2u=0:u=−54​,u=1
8−10u2+2u=0
Write in the standard form ax2+bx+c=0−10u2+2u+8=0
Solve with the quadratic formula
−10u2+2u+8=0
Quadratic Equation Formula:
For a=−10,b=2,c=8u1,2​=2(−10)−2±22−4(−10)⋅8​​
u1,2​=2(−10)−2±22−4(−10)⋅8​​
22−4(−10)⋅8​=18
22−4(−10)⋅8​
Apply rule −(−a)=a=22+4⋅10⋅8​
Multiply the numbers: 4⋅10⋅8=320=22+320​
22=4=4+320​
Add the numbers: 4+320=324=324​
Factor the number: 324=182=182​
Apply radical rule: 182​=18=18
u1,2​=2(−10)−2±18​
Separate the solutionsu1​=2(−10)−2+18​,u2​=2(−10)−2−18​
u=2(−10)−2+18​:−54​
2(−10)−2+18​
Remove parentheses: (−a)=−a=−2⋅10−2+18​
Add/Subtract the numbers: −2+18=16=−2⋅1016​
Multiply the numbers: 2⋅10=20=−2016​
Apply the fraction rule: −ba​=−ba​=−2016​
Cancel the common factor: 4=−54​
u=2(−10)−2−18​:1
2(−10)−2−18​
Remove parentheses: (−a)=−a=−2⋅10−2−18​
Subtract the numbers: −2−18=−20=−2⋅10−20​
Multiply the numbers: 2⋅10=20=−20−20​
Apply the fraction rule: −b−a​=ba​=2020​
Apply rule aa​=1=1
The solutions to the quadratic equation are:u=−54​,u=1
Substitute back u=sin(x)sin(x)=−54​,sin(x)=1
sin(x)=−54​,sin(x)=1
sin(x)=−54​:x=arcsin(−54​)+2πn,x=π+arcsin(54​)+2πn
sin(x)=−54​
Apply trig inverse properties
sin(x)=−54​
General solutions for sin(x)=−54​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−54​)+2πn,x=π+arcsin(54​)+2πn
x=arcsin(−54​)+2πn,x=π+arcsin(54​)+2πn
sin(x)=1:x=2π​+2πn
sin(x)=1
General solutions for sin(x)=1
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​​​
x=2π​+2πn
x=2π​+2πn
Combine all the solutionsx=arcsin(−54​)+2πn,x=π+arcsin(54​)+2πn,x=2π​+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into 3cos(x)+sin(x)=1
Remove the ones that don't agree with the equation.
Check the solution arcsin(−54​)+2πn:True
arcsin(−54​)+2πn
Plug in n=1arcsin(−54​)+2π1
For 3cos(x)+sin(x)=1plug inx=arcsin(−54​)+2π13cos(arcsin(−54​)+2π1)+sin(arcsin(−54​)+2π1)=1
Refine1=1
⇒True
Check the solution π+arcsin(54​)+2πn:False
π+arcsin(54​)+2πn
Plug in n=1π+arcsin(54​)+2π1
For 3cos(x)+sin(x)=1plug inx=π+arcsin(54​)+2π13cos(π+arcsin(54​)+2π1)+sin(π+arcsin(54​)+2π1)=1
Refine−2.6=1
⇒False
Check the solution 2π​+2πn:True
2π​+2πn
Plug in n=12π​+2π1
For 3cos(x)+sin(x)=1plug inx=2π​+2π13cos(2π​+2π1)+sin(2π​+2π1)=1
Refine1=1
⇒True
x=arcsin(−54​)+2πn,x=2π​+2πn
Show solutions in decimal formx=−0.92729…+2πn,x=2π​+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 3cos(x)+sin(x)=1 ?

    The general solution for 3cos(x)+sin(x)=1 is x=-0.92729…+2pin,x= pi/2+2pin
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