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Popular Trigonometry >

sin(3x)=sin^2(x)

  • Pre Algebra
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Solution

sin(3x)=sin2(x)

Solution

x=2πn,x=π+2πn,x=0.84806…+2πn,x=π−0.84806…+2πn,x=23π​+2πn
+1
Degrees
x=0∘+360∘n,x=180∘+360∘n,x=48.59037…∘+360∘n,x=131.40962…∘+360∘n,x=270∘+360∘n
Solution steps
sin(3x)=sin2(x)
Subtract sin2(x) from both sidessin(3x)−sin2(x)=0
Rewrite using trig identities
sin(3x)−sin2(x)
sin(3x)=3sin(x)−4sin3(x)
sin(3x)
Rewrite using trig identities
sin(3x)
Rewrite as=sin(2x+x)
Use the Angle Sum identity: sin(s+t)=sin(s)cos(t)+cos(s)sin(t)=sin(2x)cos(x)+cos(2x)sin(x)
Use the Double Angle identity: sin(2x)=2sin(x)cos(x)=cos(2x)sin(x)+cos(x)2sin(x)cos(x)
Simplify cos(2x)sin(x)+cos(x)⋅2sin(x)cos(x):sin(x)cos(2x)+2cos2(x)sin(x)
cos(2x)sin(x)+cos(x)2sin(x)cos(x)
cos(x)⋅2sin(x)cos(x)=2cos2(x)sin(x)
cos(x)2sin(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=2sin(x)cos1+1(x)
Add the numbers: 1+1=2=2sin(x)cos2(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
Use the Double Angle identity: cos(2x)=1−2sin2(x)=(1−2sin2(x))sin(x)+2cos2(x)sin(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=(1−2sin2(x))sin(x)+2(1−sin2(x))sin(x)
Expand (1−2sin2(x))sin(x)+2(1−sin2(x))sin(x):−4sin3(x)+3sin(x)
(1−2sin2(x))sin(x)+2(1−sin2(x))sin(x)
=sin(x)(1−2sin2(x))+2sin(x)(1−sin2(x))
Expand sin(x)(1−2sin2(x)):sin(x)−2sin3(x)
sin(x)(1−2sin2(x))
Apply the distributive law: a(b−c)=ab−aca=sin(x),b=1,c=2sin2(x)=sin(x)1−sin(x)2sin2(x)
=1sin(x)−2sin2(x)sin(x)
Simplify 1⋅sin(x)−2sin2(x)sin(x):sin(x)−2sin3(x)
1sin(x)−2sin2(x)sin(x)
1⋅sin(x)=sin(x)
1sin(x)
Multiply: 1⋅sin(x)=sin(x)=sin(x)
2sin2(x)sin(x)=2sin3(x)
2sin2(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=2sin2+1(x)
Add the numbers: 2+1=3=2sin3(x)
=sin(x)−2sin3(x)
=sin(x)−2sin3(x)
=sin(x)−2sin3(x)+2(1−sin2(x))sin(x)
Expand 2sin(x)(1−sin2(x)):2sin(x)−2sin3(x)
2sin(x)(1−sin2(x))
Apply the distributive law: a(b−c)=ab−aca=2sin(x),b=1,c=sin2(x)=2sin(x)1−2sin(x)sin2(x)
=2⋅1sin(x)−2sin2(x)sin(x)
Simplify 2⋅1⋅sin(x)−2sin2(x)sin(x):2sin(x)−2sin3(x)
2⋅1sin(x)−2sin2(x)sin(x)
2⋅1⋅sin(x)=2sin(x)
2⋅1sin(x)
Multiply the numbers: 2⋅1=2=2sin(x)
2sin2(x)sin(x)=2sin3(x)
2sin2(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=2sin2+1(x)
Add the numbers: 2+1=3=2sin3(x)
=2sin(x)−2sin3(x)
=2sin(x)−2sin3(x)
=sin(x)−2sin3(x)+2sin(x)−2sin3(x)
Simplify sin(x)−2sin3(x)+2sin(x)−2sin3(x):−4sin3(x)+3sin(x)
sin(x)−2sin3(x)+2sin(x)−2sin3(x)
Group like terms=−2sin3(x)−2sin3(x)+sin(x)+2sin(x)
Add similar elements: −2sin3(x)−2sin3(x)=−4sin3(x)=−4sin3(x)+sin(x)+2sin(x)
Add similar elements: sin(x)+2sin(x)=3sin(x)=−4sin3(x)+3sin(x)
=−4sin3(x)+3sin(x)
=−4sin3(x)+3sin(x)
=3sin(x)−4sin3(x)−sin2(x)
−sin2(x)+3sin(x)−4sin3(x)=0
Solve by substitution
−sin2(x)+3sin(x)−4sin3(x)=0
Let: sin(x)=u−u2+3u−4u3=0
−u2+3u−4u3=0:u=0,u=43​,u=−1
−u2+3u−4u3=0
Factor −u2+3u−4u3:−u(4u−3)(u+1)
−u2+3u−4u3
Factor out common term −u:−u(4u2+u−3)
−4u3−u2+3u
Apply exponent rule: ab+c=abacu2=uu=−4u2u−uu+3u
Factor out common term −u=−u(4u2+u−3)
=−u(4u2+u−3)
Factor 4u2+u−3:(4u−3)(u+1)
4u2+u−3
Write in the standard form ax2+bx+c=4u2+u−3
Break the expression into groups
4u2+u−3
Definition
Factors of 12:1,2,3,4,6,12
12
Divisors (Factors)
Find the Prime factors of 12:2,2,3
12
12divides by 212=6⋅2=2⋅6
6divides by 26=3⋅2=2⋅2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅2⋅3
Multiply the prime factors of 12:4,6
2⋅2=42⋅3=6
4,6
4,6
Add the prime factors: 2,3
Add 1 and the number 12 itself1,12
The factors of 121,2,3,4,6,12
Negative factors of 12:−1,−2,−3,−4,−6,−12
Multiply the factors by −1 to get the negative factors−1,−2,−3,−4,−6,−12
For every two factors such that u∗v=−12,check if u+v=1
Check u=1,v=−12:u∗v=−12,u+v=−11⇒FalseCheck u=2,v=−6:u∗v=−12,u+v=−4⇒False
u=4,v=−3
Group into (ax2+ux)+(vx+c)(4u2−3u)+(4u−3)
=(4u2−3u)+(4u−3)
Factor out ufrom 4u2−3u:u(4u−3)
4u2−3u
Apply exponent rule: ab+c=abacu2=uu=4uu−3u
Factor out common term u=u(4u−3)
=u(4u−3)+(4u−3)
Factor out common term 4u−3=(4u−3)(u+1)
=−u(4u−3)(u+1)
−u(4u−3)(u+1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u=0or4u−3=0oru+1=0
Solve 4u−3=0:u=43​
4u−3=0
Move 3to the right side
4u−3=0
Add 3 to both sides4u−3+3=0+3
Simplify4u=3
4u=3
Divide both sides by 4
4u=3
Divide both sides by 444u​=43​
Simplifyu=43​
u=43​
Solve u+1=0:u=−1
u+1=0
Move 1to the right side
u+1=0
Subtract 1 from both sidesu+1−1=0−1
Simplifyu=−1
u=−1
The solutions areu=0,u=43​,u=−1
Substitute back u=sin(x)sin(x)=0,sin(x)=43​,sin(x)=−1
sin(x)=0,sin(x)=43​,sin(x)=−1
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)=43​:x=arcsin(43​)+2πn,x=π−arcsin(43​)+2πn
sin(x)=43​
Apply trig inverse properties
sin(x)=43​
General solutions for sin(x)=43​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(43​)+2πn,x=π−arcsin(43​)+2πn
x=arcsin(43​)+2πn,x=π−arcsin(43​)+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
Combine all the solutionsx=2πn,x=π+2πn,x=arcsin(43​)+2πn,x=π−arcsin(43​)+2πn,x=23π​+2πn
Show solutions in decimal formx=2πn,x=π+2πn,x=0.84806…+2πn,x=π−0.84806…+2πn,x=23π​+2πn

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