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Popular Trigonometry >

sin(x)sin^3(x)-sin^5(x)sin^3(x)=0

  • Pre Algebra
  • Algebra
  • Pre Calculus
  • Calculus
  • Functions
  • Linear Algebra
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Solution

sin(x)sin3(x)−sin5(x)sin3(x)=0

Solution

x=2πn,x=π+2πn,x=2π​+2πn,x=23π​+2πn
+1
Degrees
x=0∘+360∘n,x=180∘+360∘n,x=90∘+360∘n,x=270∘+360∘n
Solution steps
sin(x)sin3(x)−sin5(x)sin3(x)=0
Solve by substitution
sin(x)sin3(x)−sin5(x)sin3(x)=0
Let: sin(x)=uuu3−u5u3=0
uu3−u5u3=0:u=0,u=1,u=−1,u=i,u=−i
uu3−u5u3=0
Refineu4−u8=0
Write in the standard form an​xn+…+a1​x+a0​=0−u8+u4=0
Rewrite the equation with v=u2,v2=u4 and v4=u8−v4+v2=0
Solve −v4+v2=0:v=0,v=1,v=−1
−v4+v2=0
Rewrite the equation with u=v2 and u2=v4−u2+u=0
Solve −u2+u=0:u=0,u=1
−u2+u=0
Solve with the quadratic formula
−u2+u=0
Quadratic Equation Formula:
For a=−1,b=1,c=0u1,2​=2(−1)−1±12−4(−1)⋅0​​
u1,2​=2(−1)−1±12−4(−1)⋅0​​
12−4(−1)⋅0​=1
12−4(−1)⋅0​
Apply rule 1a=112=1=1−4(−1)⋅0​
Apply rule −(−a)=a=1+4⋅1⋅0​
Apply rule 0⋅a=0=1+0​
Add the numbers: 1+0=1=1​
Apply rule 1​=1=1
u1,2​=2(−1)−1±1​
Separate the solutionsu1​=2(−1)−1+1​,u2​=2(−1)−1−1​
u=2(−1)−1+1​:0
2(−1)−1+1​
Remove parentheses: (−a)=−a=−2⋅1−1+1​
Add/Subtract the numbers: −1+1=0=−2⋅10​
Multiply the numbers: 2⋅1=2=−20​
Apply the fraction rule: −ba​=−ba​=−20​
Apply rule a0​=0,a=0=−0
=0
u=2(−1)−1−1​:1
2(−1)−1−1​
Remove parentheses: (−a)=−a=−2⋅1−1−1​
Subtract the numbers: −1−1=−2=−2⋅1−2​
Multiply the numbers: 2⋅1=2=−2−2​
Apply the fraction rule: −b−a​=ba​=22​
Apply rule aa​=1=1
The solutions to the quadratic equation are:u=0,u=1
u=0,u=1
Substitute back u=v2,solve for v
Solve v2=0:v=0
v2=0
Apply rule xn=0⇒x=0
v=0
Solve v2=1:v=1,v=−1
v2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
v=1​,v=−1​
1​=1
1​
Apply rule 1​=1=1
−1​=−1
−1​
Apply rule 1​=1=−1
v=1,v=−1
The solutions are
v=0,v=1,v=−1
v=0,v=1,v=−1
Substitute back v=u2,solve for u
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
Solve u2=1:u=1,u=−1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply rule 1​=1=1
−1​=−1
−1​
Apply rule 1​=1=−1
u=1,u=−1
Solve u2=−1:u=i,u=−i
u2=−1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=−1​,u=−−1​
Simplify −1​:i
−1​
Apply imaginary number rule: −1​=i=i
Simplify −−1​:−i
−−1​
Apply imaginary number rule: −1​=i=−i
u=i,u=−i
The solutions are
u=0,u=1,u=−1,u=i,u=−i
Substitute back u=sin(x)sin(x)=0,sin(x)=1,sin(x)=−1,sin(x)=i,sin(x)=−i
sin(x)=0,sin(x)=1,sin(x)=−1,sin(x)=i,sin(x)=−i
sin(x)=0:x=2πn,x=π+2πn
sin(x)=0
General solutions for sin(x)=0
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=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+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
sin(x)=−1:x=23π​+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=23π​+2πn
x=23π​+2πn
sin(x)=i:No Solution
sin(x)=i
NoSolution
sin(x)=−i:No Solution
sin(x)=−i
NoSolution
Combine all the solutionsx=2πn,x=π+2πn,x=2π​+2πn,x=23π​+2πn

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