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Popular Trigonometry >

tan(1/2 x)=3cos(1/2 x)

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Solution

tan(21​x)=3cos(21​x)

Solution

x=2⋅1.01055…+4πn,x=2π−2⋅1.01055…+4πn
+1
Degrees
x=115.80111…∘+720∘n,x=244.19888…∘+720∘n
Solution steps
tan(21​x)=3cos(21​x)
Subtract 3cos(21​x) from both sidestan(2x​)−3cos(2x​)=0
Express with sin, coscos(2x​)sin(2x​)​−3cos(2x​)=0
Simplify cos(2x​)sin(2x​)​−3cos(2x​):cos(2x​)sin(2x​)−3cos2(2x​)​
cos(2x​)sin(2x​)​−3cos(2x​)
Convert element to fraction: 3cos(2x​)=cos(2x​)3cos(2x​)cos(2x​)​=cos(2x​)sin(2x​)​−cos(2x​)3cos(2x​)cos(2x​)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=cos(2x​)sin(2x​)−3cos(2x​)cos(2x​)​
sin(2x​)−3cos(2x​)cos(2x​)=sin(2x​)−3cos2(2x​)
sin(2x​)−3cos(2x​)cos(2x​)
3cos(2x​)cos(2x​)=3cos2(2x​)
3cos(2x​)cos(2x​)
Apply exponent rule: ab⋅ac=ab+ccos(2x​)cos(2x​)=cos1+1(2x​)=3cos1+1(2x​)
Add the numbers: 1+1=2=3cos2(2x​)
=sin(2x​)−3cos2(2x​)
=cos(2x​)sin(2x​)−3cos2(2x​)​
cos(2x​)sin(2x​)−3cos2(2x​)​=0
g(x)f(x)​=0⇒f(x)=0sin(2x​)−3cos2(2x​)=0
Add 3cos2(2x​) to both sidessin(2x​)=3cos2(2x​)
Square both sidessin2(2x​)=(3cos2(2x​))2
Subtract (3cos2(2x​))2 from both sidessin2(2x​)−9cos4(2x​)=0
Factor sin2(2x​)−9cos4(2x​):(sin(2x​)+3cos2(2x​))(sin(2x​)−3cos2(2x​))
sin2(2x​)−9cos4(2x​)
Rewrite sin2(2x​)−9cos4(2x​) as sin2(2x​)−(3cos2(2x​))2
sin2(2x​)−9cos4(2x​)
Rewrite 9 as 32=sin2(2x​)−32cos4(2x​)
Apply exponent rule: abc=(ab)ccos4(2x​)=(cos2(2x​))2=sin2(2x​)−32(cos2(2x​))2
Apply exponent rule: ambm=(ab)m32(cos2(2x​))2=(3cos2(2x​))2=sin2(2x​)−(3cos2(2x​))2
=sin2(2x​)−(3cos2(2x​))2
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)sin2(2x​)−(3cos2(2x​))2=(sin(2x​)+3cos2(2x​))(sin(2x​)−3cos2(2x​))=(sin(2x​)+3cos2(2x​))(sin(2x​)−3cos2(2x​))
(sin(2x​)+3cos2(2x​))(sin(2x​)−3cos2(2x​))=0
Solving each part separatelysin(2x​)+3cos2(2x​)=0orsin(2x​)−3cos2(2x​)=0
sin(2x​)+3cos2(2x​)=0:x=−2arcsin(637​−1​)+4πn,x=2π+2arcsin(6−1+37​​)+4πn
sin(2x​)+3cos2(2x​)=0
Rewrite using trig identities
sin(2x​)+3cos2(2x​)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=sin(2x​)+3(1−sin2(2x​))
sin(2x​)+(1−sin2(2x​))⋅3=0
Solve by substitution
sin(2x​)+(1−sin2(2x​))⋅3=0
Let: sin(2x​)=uu+(1−u2)⋅3=0
u+(1−u2)⋅3=0:u=−6−1+37​​,u=61+37​​
u+(1−u2)⋅3=0
Expand u+(1−u2)⋅3:u+3−3u2
u+(1−u2)⋅3
=u+3(1−u2)
Expand 3(1−u2):3−3u2
3(1−u2)
Apply the distributive law: a(b−c)=ab−aca=3,b=1,c=u2=3⋅1−3u2
Multiply the numbers: 3⋅1=3=3−3u2
=u+3−3u2
u+3−3u2=0
Write in the standard form ax2+bx+c=0−3u2+u+3=0
Solve with the quadratic formula
−3u2+u+3=0
Quadratic Equation Formula:
For a=−3,b=1,c=3u1,2​=2(−3)−1±12−4(−3)⋅3​​
u1,2​=2(−3)−1±12−4(−3)⋅3​​
12−4(−3)⋅3​=37​
12−4(−3)⋅3​
Apply rule 1a=112=1=1−4(−3)⋅3​
Apply rule −(−a)=a=1+4⋅3⋅3​
Multiply the numbers: 4⋅3⋅3=36=1+36​
Add the numbers: 1+36=37=37​
u1,2​=2(−3)−1±37​​
Separate the solutionsu1​=2(−3)−1+37​​,u2​=2(−3)−1−37​​
u=2(−3)−1+37​​:−6−1+37​​
2(−3)−1+37​​
Remove parentheses: (−a)=−a=−2⋅3−1+37​​
Multiply the numbers: 2⋅3=6=−6−1+37​​
Apply the fraction rule: −ba​=−ba​=−6−1+37​​
u=2(−3)−1−37​​:61+37​​
2(−3)−1−37​​
Remove parentheses: (−a)=−a=−2⋅3−1−37​​
Multiply the numbers: 2⋅3=6=−6−1−37​​
Apply the fraction rule: −b−a​=ba​−1−37​=−(1+37​)=61+37​​
The solutions to the quadratic equation are:u=−6−1+37​​,u=61+37​​
Substitute back u=sin(2x​)sin(2x​)=−6−1+37​​,sin(2x​)=61+37​​
sin(2x​)=−6−1+37​​,sin(2x​)=61+37​​
sin(2x​)=−6−1+37​​:x=−2arcsin(637​−1​)+4πn,x=2π+2arcsin(6−1+37​​)+4πn
sin(2x​)=−6−1+37​​
Apply trig inverse properties
sin(2x​)=−6−1+37​​
General solutions for sin(2x​)=−6−1+37​​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πn2x​=arcsin(−6−1+37​​)+2πn,2x​=π+arcsin(6−1+37​​)+2πn
2x​=arcsin(−6−1+37​​)+2πn,2x​=π+arcsin(6−1+37​​)+2πn
Solve 2x​=arcsin(−6−1+37​​)+2πn:x=−2arcsin(637​−1​)+4πn
2x​=arcsin(−6−1+37​​)+2πn
Simplify arcsin(−6−1+37​​)+2πn:−arcsin(637​−1​)+2πn
arcsin(−6−1+37​​)+2πn
Use the following property: arcsin(−x)=−arcsin(x)arcsin(−637​−1​)=−arcsin(637​−1​)=−arcsin(637​−1​)+2πn
2x​=−arcsin(637​−1​)+2πn
Multiply both sides by 2
2x​=−arcsin(637​−1​)+2πn
Multiply both sides by 222x​=−2arcsin(637​−1​)+2⋅2πn
Simplify
22x​=−2arcsin(637​−1​)+2⋅2πn
Simplify 22x​:x
22x​
Divide the numbers: 22​=1=x
Simplify −2arcsin(637​−1​)+2⋅2πn:−2arcsin(637​−1​)+4πn
−2arcsin(637​−1​)+2⋅2πn
Multiply the numbers: 2⋅2=4=−2arcsin(637​−1​)+4πn
x=−2arcsin(637​−1​)+4πn
x=−2arcsin(637​−1​)+4πn
x=−2arcsin(637​−1​)+4πn
Solve 2x​=π+arcsin(6−1+37​​)+2πn:x=2π+2arcsin(6−1+37​​)+4πn
2x​=π+arcsin(6−1+37​​)+2πn
Multiply both sides by 2
2x​=π+arcsin(6−1+37​​)+2πn
Multiply both sides by 222x​=2π+2arcsin(6−1+37​​)+2⋅2πn
Simplify
22x​=2π+2arcsin(6−1+37​​)+2⋅2πn
Simplify 22x​:x
22x​
Divide the numbers: 22​=1=x
Simplify 2π+2arcsin(6−1+37​​)+2⋅2πn:2π+2arcsin(6−1+37​​)+4πn
2π+2arcsin(6−1+37​​)+2⋅2πn
Multiply the numbers: 2⋅2=4=2π+2arcsin(637​−1​)+4πn
x=2π+2arcsin(6−1+37​​)+4πn
x=2π+2arcsin(6−1+37​​)+4πn
x=2π+2arcsin(6−1+37​​)+4πn
x=−2arcsin(637​−1​)+4πn,x=2π+2arcsin(6−1+37​​)+4πn
sin(2x​)=61+37​​:No Solution
sin(2x​)=61+37​​
−1≤sin(x)≤1NoSolution
Combine all the solutionsx=−2arcsin(637​−1​)+4πn,x=2π+2arcsin(6−1+37​​)+4πn
sin(2x​)−3cos2(2x​)=0:x=2arcsin(6−1+37​​)+4πn,x=2π−2arcsin(6−1+37​​)+4πn
sin(2x​)−3cos2(2x​)=0
Rewrite using trig identities
sin(2x​)−3cos2(2x​)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=sin(2x​)−3(1−sin2(2x​))
sin(2x​)−(1−sin2(2x​))⋅3=0
Solve by substitution
sin(2x​)−(1−sin2(2x​))⋅3=0
Let: sin(2x​)=uu−(1−u2)⋅3=0
u−(1−u2)⋅3=0:u=6−1+37​​,u=6−1−37​​
u−(1−u2)⋅3=0
Expand u−(1−u2)⋅3:u−3+3u2
u−(1−u2)⋅3
=u−3(1−u2)
Expand −3(1−u2):−3+3u2
−3(1−u2)
Apply the distributive law: a(b−c)=ab−aca=−3,b=1,c=u2=−3⋅1−(−3)u2
Apply minus-plus rules−(−a)=a=−3⋅1+3u2
Multiply the numbers: 3⋅1=3=−3+3u2
=u−3+3u2
u−3+3u2=0
Write in the standard form ax2+bx+c=03u2+u−3=0
Solve with the quadratic formula
3u2+u−3=0
Quadratic Equation Formula:
For a=3,b=1,c=−3u1,2​=2⋅3−1±12−4⋅3(−3)​​
u1,2​=2⋅3−1±12−4⋅3(−3)​​
12−4⋅3(−3)​=37​
12−4⋅3(−3)​
Apply rule 1a=112=1=1−4⋅3(−3)​
Apply rule −(−a)=a=1+4⋅3⋅3​
Multiply the numbers: 4⋅3⋅3=36=1+36​
Add the numbers: 1+36=37=37​
u1,2​=2⋅3−1±37​​
Separate the solutionsu1​=2⋅3−1+37​​,u2​=2⋅3−1−37​​
u=2⋅3−1+37​​:6−1+37​​
2⋅3−1+37​​
Multiply the numbers: 2⋅3=6=6−1+37​​
u=2⋅3−1−37​​:6−1−37​​
2⋅3−1−37​​
Multiply the numbers: 2⋅3=6=6−1−37​​
The solutions to the quadratic equation are:u=6−1+37​​,u=6−1−37​​
Substitute back u=sin(2x​)sin(2x​)=6−1+37​​,sin(2x​)=6−1−37​​
sin(2x​)=6−1+37​​,sin(2x​)=6−1−37​​
sin(2x​)=6−1+37​​:x=2arcsin(6−1+37​​)+4πn,x=2π−2arcsin(6−1+37​​)+4πn
sin(2x​)=6−1+37​​
Apply trig inverse properties
sin(2x​)=6−1+37​​
General solutions for sin(2x​)=6−1+37​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πn2x​=arcsin(6−1+37​​)+2πn,2x​=π−arcsin(6−1+37​​)+2πn
2x​=arcsin(6−1+37​​)+2πn,2x​=π−arcsin(6−1+37​​)+2πn
Solve 2x​=arcsin(6−1+37​​)+2πn:x=2arcsin(6−1+37​​)+4πn
2x​=arcsin(6−1+37​​)+2πn
Multiply both sides by 2
2x​=arcsin(6−1+37​​)+2πn
Multiply both sides by 222x​=2arcsin(6−1+37​​)+2⋅2πn
Simplify
22x​=2arcsin(6−1+37​​)+2⋅2πn
Simplify 22x​:x
22x​
Divide the numbers: 22​=1=x
Simplify 2arcsin(6−1+37​​)+2⋅2πn:2arcsin(6−1+37​​)+4πn
2arcsin(6−1+37​​)+2⋅2πn
Multiply the numbers: 2⋅2=4=2arcsin(637​−1​)+4πn
x=2arcsin(6−1+37​​)+4πn
x=2arcsin(6−1+37​​)+4πn
x=2arcsin(6−1+37​​)+4πn
Solve 2x​=π−arcsin(6−1+37​​)+2πn:x=2π−2arcsin(6−1+37​​)+4πn
2x​=π−arcsin(6−1+37​​)+2πn
Multiply both sides by 2
2x​=π−arcsin(6−1+37​​)+2πn
Multiply both sides by 222x​=2π−2arcsin(6−1+37​​)+2⋅2πn
Simplify
22x​=2π−2arcsin(6−1+37​​)+2⋅2πn
Simplify 22x​:x
22x​
Divide the numbers: 22​=1=x
Simplify 2π−2arcsin(6−1+37​​)+2⋅2πn:2π−2arcsin(6−1+37​​)+4πn
2π−2arcsin(6−1+37​​)+2⋅2πn
Multiply the numbers: 2⋅2=4=2π−2arcsin(637​−1​)+4πn
x=2π−2arcsin(6−1+37​​)+4πn
x=2π−2arcsin(6−1+37​​)+4πn
x=2π−2arcsin(6−1+37​​)+4πn
x=2arcsin(6−1+37​​)+4πn,x=2π−2arcsin(6−1+37​​)+4πn
sin(2x​)=6−1−37​​:No Solution
sin(2x​)=6−1−37​​
−1≤sin(x)≤1NoSolution
Combine all the solutionsx=2arcsin(6−1+37​​)+4πn,x=2π−2arcsin(6−1+37​​)+4πn
Combine all the solutionsx=−2arcsin(637​−1​)+4πn,x=2π+2arcsin(6−1+37​​)+4πn,x=2arcsin(6−1+37​​)+4πn,x=2π−2arcsin(6−1+37​​)+4πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into tan(21​x)=3cos(21​x)
Remove the ones that don't agree with the equation.
Check the solution −2arcsin(637​−1​)+4πn:False
−2arcsin(637​−1​)+4πn
Plug in n=1−2arcsin(637​−1​)+4π1
For tan(21​x)=3cos(21​x)plug inx=−2arcsin(637​−1​)+4π1tan(21​(−2arcsin(637​−1​)+4π1))=3cos(21​(−2arcsin(637​−1​)+4π1))
Refine−1.59417…=1.59417…
⇒False
Check the solution 2π+2arcsin(6−1+37​​)+4πn:False
2π+2arcsin(6−1+37​​)+4πn
Plug in n=12π+2arcsin(6−1+37​​)+4π1
For tan(21​x)=3cos(21​x)plug inx=2π+2arcsin(6−1+37​​)+4π1tan(21​(2π+2arcsin(6−1+37​​)+4π1))=3cos(21​(2π+2arcsin(6−1+37​​)+4π1))
Refine1.59417…=−1.59417…
⇒False
Check the solution 2arcsin(6−1+37​​)+4πn:True
2arcsin(6−1+37​​)+4πn
Plug in n=12arcsin(6−1+37​​)+4π1
For tan(21​x)=3cos(21​x)plug inx=2arcsin(6−1+37​​)+4π1tan(21​(2arcsin(6−1+37​​)+4π1))=3cos(21​(2arcsin(6−1+37​​)+4π1))
Refine1.59417…=1.59417…
⇒True
Check the solution 2π−2arcsin(6−1+37​​)+4πn:True
2π−2arcsin(6−1+37​​)+4πn
Plug in n=12π−2arcsin(6−1+37​​)+4π1
For tan(21​x)=3cos(21​x)plug inx=2π−2arcsin(6−1+37​​)+4π1tan(21​(2π−2arcsin(6−1+37​​)+4π1))=3cos(21​(2π−2arcsin(6−1+37​​)+4π1))
Refine−1.59417…=−1.59417…
⇒True
x=2arcsin(6−1+37​​)+4πn,x=2π−2arcsin(6−1+37​​)+4πn
Show solutions in decimal formx=2⋅1.01055…+4πn,x=2π−2⋅1.01055…+4πn

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Popular Examples

cos(2θ)=-1/2 ,0<= x<= 2picos(x)=-5/13sin^2(x)=6(cos(x)+1)2cos^2(x)-cos(x)=1,0<= x<= 2pi7sin^2(θ)-36sin(θ)+5=0

Frequently Asked Questions (FAQ)

  • What is the general solution for tan(1/2 x)=3cos(1/2 x) ?

    The general solution for tan(1/2 x)=3cos(1/2 x) is x=2*1.01055…+4pin,x=2pi-2*1.01055…+4pin
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