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

Solution

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

Solution

x = 2 \cdot 1.01055 \dots + 4 πn, x = 2 π - 2 \cdot 1.01055 \dots + 4 πn

Degrees

x = 115.80111 \dots^{\circ} + 72 0^{\circ} n, x = 244.19888 \dots^{\circ} + 72 0^{\circ} n

Solution steps

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

Subtract

3 cos (\frac{1}{2} x)

from both sides

tan (\frac{x}{2}) - 3 cos (\frac{x}{2}) = 0

Express with sin, cos

\frac{sin ( \frac{x}{2} )}{cos ( \frac{x}{2} )} - 3 cos (\frac{x}{2}) = 0

Simplify

\frac{sin ( \frac{x}{2} )}{cos ( \frac{x}{2} )} - 3 cos (\frac{x}{2}) : \frac{sin ( \frac{x}{2} ) - 3 cos ^{2} ( \frac{x}{2} )}{cos ( \frac{x}{2} )}

\frac{sin ( \frac{x}{2} )}{cos ( \frac{x}{2} )} - 3 cos (\frac{x}{2})

Convert element to fraction:

3 cos (\frac{x}{2}) = \frac{3 cos ( \frac{x}{2} ) cos ( \frac{x}{2} )}{cos ( \frac{x}{2} )}

= \frac{sin ( \frac{x}{2} )}{cos ( \frac{x}{2} )} - \frac{3 cos ( \frac{x}{2} ) cos ( \frac{x}{2} )}{cos ( \frac{x}{2} )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( \frac{x}{2} ) - 3 cos ( \frac{x}{2} ) cos ( \frac{x}{2} )}{cos ( \frac{x}{2} )}

sin (\frac{x}{2}) - 3 cos (\frac{x}{2}) cos (\frac{x}{2}) = sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2})

sin (\frac{x}{2}) - 3 cos (\frac{x}{2}) cos (\frac{x}{2})

3 cos (\frac{x}{2}) cos (\frac{x}{2}) = 3 cos^{2} (\frac{x}{2})

3 cos (\frac{x}{2}) cos (\frac{x}{2})

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (\frac{x}{2}) cos (\frac{x}{2}) = cos^{1 + 1} (\frac{x}{2})

= 3 cos^{1 + 1} (\frac{x}{2})

Add the numbers:

1 + 1 = 2

= 3 cos^{2} (\frac{x}{2})

= sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2})

= \frac{sin ( \frac{x}{2} ) - 3 cos ^{2} ( \frac{x}{2} )}{cos ( \frac{x}{2} )}

\frac{sin ( \frac{x}{2} ) - 3 cos ^{2} ( \frac{x}{2} )}{cos ( \frac{x}{2} )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}) = 0

Add

3 cos^{2} (\frac{x}{2})

to both sides

sin (\frac{x}{2}) = 3 cos^{2} (\frac{x}{2})

Square both sides

sin^{2} (\frac{x}{2}) = (3 cos^{2} (\frac{x}{2}))^{2}

Subtract

(3 cos^{2} (\frac{x}{2}))^{2}

from both sides

sin^{2} (\frac{x}{2}) - 9 cos^{4} (\frac{x}{2}) = 0

Factor

sin^{2} (\frac{x}{2}) - 9 cos^{4} (\frac{x}{2}) : (sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2})) (sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}))

sin^{2} (\frac{x}{2}) - 9 cos^{4} (\frac{x}{2})

Rewrite

sin^{2} (\frac{x}{2}) - 9 cos^{4} (\frac{x}{2})

sin^{2} (\frac{x}{2}) - (3 cos^{2} (\frac{x}{2}))^{2}

sin^{2} (\frac{x}{2}) - 9 cos^{4} (\frac{x}{2})

Rewrite

9

3^{2}

= sin^{2} (\frac{x}{2}) - 3^{2} cos^{4} (\frac{x}{2})

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (\frac{x}{2}) = (cos^{2} (\frac{x}{2}))^{2}

= sin^{2} (\frac{x}{2}) - 3^{2} (cos^{2} (\frac{x}{2}))^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

3^{2} (cos^{2} (\frac{x}{2}))^{2} = (3 cos^{2} (\frac{x}{2}))^{2}

= sin^{2} (\frac{x}{2}) - (3 cos^{2} (\frac{x}{2}))^{2}

= sin^{2} (\frac{x}{2}) - (3 cos^{2} (\frac{x}{2}))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sin^{2} (\frac{x}{2}) - (3 cos^{2} (\frac{x}{2}))^{2} = (sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2})) (sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}))

= (sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2})) (sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}))

(sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2})) (sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2})) = 0

Solving each part separately

sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2}) = 0 or sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}) = 0

sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2}) = 0 : x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn, x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2}) = 0

Rewrite using trig identities

sin (\frac{x}{2}) + 3 cos^{2} (\frac{x}{2})

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= sin (\frac{x}{2}) + 3 (1 - sin^{2} (\frac{x}{2}))

sin (\frac{x}{2}) + (1 - sin^{2} (\frac{x}{2})) \cdot 3 = 0

Solve by substitution

sin (\frac{x}{2}) + (1 - sin^{2} (\frac{x}{2})) \cdot 3 = 0

Let:

sin (\frac{x}{2}) = u

u + (1 - u^{2}) \cdot 3 = 0

u + (1 - u^{2}) \cdot 3 = 0 : u = - \frac{- 1 + 37}{6}, u = \frac{1 + 37}{6}

u + (1 - u^{2}) \cdot 3 = 0

Expand

u + (1 - u^{2}) \cdot 3 : u + 3 - 3 u^{2}

u + (1 - u^{2}) \cdot 3

= u + 3 (1 - u^{2})

Expand

3 (1 - u^{2}) : 3 - 3 u^{2}

3 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = u^{2}

= 3 \cdot 1 - 3 u^{2}

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 u^{2}

= u + 3 - 3 u^{2}

u + 3 - 3 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 3 u^{2} + u + 3 = 0

Solve with the quadratic formula

- 3 u^{2} + u + 3 = 0

Quadratic Equation Formula:

For

a = - 3, b = 1, c = 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 3}{2 ( - 3 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 3}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 3 = 37

1^{2} - 4 (- 3) \cdot 3

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 3) \cdot 3

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 + 36

Add the numbers:

1 + 36 = 37

= 37

u_{1, 2} = \frac{- 1 \pm 37}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 1 + 37}{2 ( - 3 )}, u_{2} = \frac{- 1 - 37}{2 ( - 3 )}

u = \frac{- 1 + 37}{2 ( - 3 )} : - \frac{- 1 + 37}{6}

\frac{- 1 + 37}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 37}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 1 + 37}{- 6}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 37}{6}

u = \frac{- 1 - 37}{2 ( - 3 )} : \frac{1 + 37}{6}

\frac{- 1 - 37}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 37}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 1 - 37}{- 6}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

- 1 - 37 = - (1 + 37)

= \frac{1 + 37}{6}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 37}{6}, u = \frac{1 + 37}{6}

Substitute back

u = sin (\frac{x}{2})

sin (\frac{x}{2}) = - \frac{- 1 + 37}{6}, sin (\frac{x}{2}) = \frac{1 + 37}{6}

sin (\frac{x}{2}) = - \frac{- 1 + 37}{6}, sin (\frac{x}{2}) = \frac{1 + 37}{6}

sin (\frac{x}{2}) = - \frac{- 1 + 37}{6} : x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn, x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) = - \frac{- 1 + 37}{6}

Apply trig inverse properties

sin (\frac{x}{2}) = - \frac{- 1 + 37}{6}

General solutions for

sin (\frac{x}{2}) = - \frac{- 1 + 37}{6}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

\frac{x}{2} = arcsin (- \frac{- 1 + 37}{6}) + 2 πn, \frac{x}{2} = π + arcsin (\frac{- 1 + 37}{6}) + 2 πn

\frac{x}{2} = arcsin (- \frac{- 1 + 37}{6}) + 2 πn, \frac{x}{2} = π + arcsin (\frac{- 1 + 37}{6}) + 2 πn

Solve

\frac{x}{2} = arcsin (- \frac{- 1 + 37}{6}) + 2 πn : x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

\frac{x}{2} = arcsin (- \frac{- 1 + 37}{6}) + 2 πn

Simplify

arcsin (- \frac{- 1 + 37}{6}) + 2 πn : - arcsin (\frac{37 - 1}{6}) + 2 πn

arcsin (- \frac{- 1 + 37}{6}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{37 - 1}{6}) = - arcsin (\frac{37 - 1}{6})

= - arcsin (\frac{37 - 1}{6}) + 2 πn

\frac{x}{2} = - arcsin (\frac{37 - 1}{6}) + 2 πn

Multiply both sides by

2

\frac{x}{2} = - arcsin (\frac{37 - 1}{6}) + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = - 2 arcsin (\frac{37 - 1}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = - 2 arcsin (\frac{37 - 1}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- 2 arcsin (\frac{37 - 1}{6}) + 2 \cdot 2 πn : - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

- 2 arcsin (\frac{37 - 1}{6}) + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

Solve

\frac{x}{2} = π + arcsin (\frac{- 1 + 37}{6}) + 2 πn : x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

\frac{x}{2} = π + arcsin (\frac{- 1 + 37}{6}) + 2 πn

Multiply both sides by

2

\frac{x}{2} = π + arcsin (\frac{- 1 + 37}{6}) + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn : 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 2 π + 2 arcsin (\frac{37 - 1}{6}) + 4 πn

x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn, x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) = \frac{1 + 37}{6} :

No Solution

sin (\frac{x}{2}) = \frac{1 + 37}{6}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn, x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}) = 0 : x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2}) = 0

Rewrite using trig identities

sin (\frac{x}{2}) - 3 cos^{2} (\frac{x}{2})

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= sin (\frac{x}{2}) - 3 (1 - sin^{2} (\frac{x}{2}))

sin (\frac{x}{2}) - (1 - sin^{2} (\frac{x}{2})) \cdot 3 = 0

Solve by substitution

sin (\frac{x}{2}) - (1 - sin^{2} (\frac{x}{2})) \cdot 3 = 0

Let:

sin (\frac{x}{2}) = u

u - (1 - u^{2}) \cdot 3 = 0

u - (1 - u^{2}) \cdot 3 = 0 : u = \frac{- 1 + 37}{6}, u = \frac{- 1 - 37}{6}

u - (1 - u^{2}) \cdot 3 = 0

Expand

u - (1 - u^{2}) \cdot 3 : u - 3 + 3 u^{2}

u - (1 - u^{2}) \cdot 3

= u - 3 (1 - u^{2})

Expand

- 3 (1 - u^{2}) : - 3 + 3 u^{2}

- 3 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = u^{2}

= - 3 \cdot 1 - (- 3) u^{2}

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 u^{2}

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 u^{2}

= u - 3 + 3 u^{2}

u - 3 + 3 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

3 u^{2} + u - 3 = 0

Solve with the quadratic formula

3 u^{2} + u - 3 = 0

Quadratic Equation Formula:

For

a = 3, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 3 )}{2 \cdot 3}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 3 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 3) = 37

1^{2} - 4 \cdot 3 (- 3)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 + 36

Add the numbers:

1 + 36 = 37

= 37

u_{1, 2} = \frac{- 1 \pm 37}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 1 + 37}{2 \cdot 3}, u_{2} = \frac{- 1 - 37}{2 \cdot 3}

u = \frac{- 1 + 37}{2 \cdot 3} : \frac{- 1 + 37}{6}

\frac{- 1 + 37}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 1 + 37}{6}

u = \frac{- 1 - 37}{2 \cdot 3} : \frac{- 1 - 37}{6}

\frac{- 1 - 37}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 1 - 37}{6}

The solutions to the quadratic equation are:

u = \frac{- 1 + 37}{6}, u = \frac{- 1 - 37}{6}

Substitute back

u = sin (\frac{x}{2})

sin (\frac{x}{2}) = \frac{- 1 + 37}{6}, sin (\frac{x}{2}) = \frac{- 1 - 37}{6}

sin (\frac{x}{2}) = \frac{- 1 + 37}{6}, sin (\frac{x}{2}) = \frac{- 1 - 37}{6}

sin (\frac{x}{2}) = \frac{- 1 + 37}{6} : x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) = \frac{- 1 + 37}{6}

Apply trig inverse properties

sin (\frac{x}{2}) = \frac{- 1 + 37}{6}

General solutions for

sin (\frac{x}{2}) = \frac{- 1 + 37}{6}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

\frac{x}{2} = arcsin (\frac{- 1 + 37}{6}) + 2 πn, \frac{x}{2} = π - arcsin (\frac{- 1 + 37}{6}) + 2 πn

\frac{x}{2} = arcsin (\frac{- 1 + 37}{6}) + 2 πn, \frac{x}{2} = π - arcsin (\frac{- 1 + 37}{6}) + 2 πn

Solve

\frac{x}{2} = arcsin (\frac{- 1 + 37}{6}) + 2 πn : x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

\frac{x}{2} = arcsin (\frac{- 1 + 37}{6}) + 2 πn

Multiply both sides by

2

\frac{x}{2} = arcsin (\frac{- 1 + 37}{6}) + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn : 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 2 arcsin (\frac{37 - 1}{6}) + 4 πn

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Solve

\frac{x}{2} = π - arcsin (\frac{- 1 + 37}{6}) + 2 πn : x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

\frac{x}{2} = π - arcsin (\frac{- 1 + 37}{6}) + 2 πn

Multiply both sides by

2

\frac{x}{2} = π - arcsin (\frac{- 1 + 37}{6}) + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn : 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 2 π - 2 arcsin (\frac{37 - 1}{6}) + 4 πn

x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

sin (\frac{x}{2}) = \frac{- 1 - 37}{6} :

No Solution

sin (\frac{x}{2}) = \frac{- 1 - 37}{6}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Combine all the solutions

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 πn, x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

Remove the ones that don't agree with the equation.

Check the solution

- 2 arcsin (\frac{37 - 1}{6}) + 4 πn :

False

- 2 arcsin (\frac{37 - 1}{6}) + 4 πn

Plug in

n = 1

- 2 arcsin (\frac{37 - 1}{6}) + 4 π 1

For

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

plug in

x = - 2 arcsin (\frac{37 - 1}{6}) + 4 π 1

tan (\frac{1}{2} (- 2 arcsin (\frac{37 - 1}{6}) + 4 π 1)) = 3 cos (\frac{1}{2} (- 2 arcsin (\frac{37 - 1}{6}) + 4 π 1))

Refine

- 1.59417 \dots = 1.59417 \dots

\Rightarrow False

Check the solution

2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn :

False

2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Plug in

n = 1

2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1

For

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

plug in

x = 2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1

tan (\frac{1}{2} (2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1)) = 3 cos (\frac{1}{2} (2 π + 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1))

Refine

1.59417 \dots = - 1.59417 \dots

\Rightarrow False

Check the solution

2 arcsin (\frac{- 1 + 37}{6}) + 4 πn :

True

2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Plug in

n = 1

2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1

For

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

plug in

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1

tan (\frac{1}{2} (2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1)) = 3 cos (\frac{1}{2} (2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1))

Refine

1.59417 \dots = 1.59417 \dots

\Rightarrow True

Check the solution

2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn :

True

2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Plug in

n = 1

2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1

For

tan (\frac{1}{2} x) = 3 cos (\frac{1}{2} x)

plug in

x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1

tan (\frac{1}{2} (2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1)) = 3 cos (\frac{1}{2} (2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 π 1))

Refine

- 1.59417 \dots = - 1.59417 \dots

\Rightarrow True

x = 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 1 + 37}{6}) + 4 πn

Show solutions in decimal form

x = 2 \cdot 1.01055 \dots + 4 πn, x = 2 π - 2 \cdot 1.01055 \dots + 4 πn

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cos (x) = - \frac{5}{13}

sin^2(x)=6(cos(x)+1)

sin^{2} (x) = 6 (cos (x) + 1)

2cos^2(x)-cos(x)=1,0<= x<= 2pi

2 cos^{2} (x) - cos (x) = 1, 0 \leq x \leq 2 π

7sin^2(θ)-36sin(θ)+5=0

7 sin^{2} (θ) - 36 sin (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