What is the general solution for tan(2x)=cos(2x),0<= x<= pi ?

Q: What is the general solution for tan(2x)=cos(2x),0<= x<= pi ?

The general solution for tan(2x)=cos(2x),0<= x<= pi is x=(0.66623…)/2 ,x=(pi-0.66623…)/2

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Popular Trigonometry >

tan(2x)=cos(2x),0<= x<= pi

Solution

tan (2 x) = cos (2 x), 0 \leq x \leq π

Solution

x = \frac{0.66623 \dots}{2}, x = \frac{π - 0.66623 \dots}{2}

Degrees

x = 19.08635 \dots^{\circ}, x = 70.91364 \dots^{\circ}

Solution steps

tan (2 x) = cos (2 x), 0 \leq x \leq π

Subtract

cos (2 x)

from both sides

tan (2 x) - cos (2 x) = 0

Express with sin, cos

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (2 x) - cos (2 x) cos (2 x)

cos (2 x) cos (2 x) = cos^{2} (2 x)

cos (2 x) cos (2 x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= cos^{2} (2 x)

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

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

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

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

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

Add

cos^{2} (2 x)

to both sides

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

Square both sides

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

Subtract

(cos^{2} (2 x))^{2}

from both sides

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

Factor

sin^{2} (2 x) - cos^{4} (2 x) : (sin (2 x) + cos^{2} (2 x)) (sin (2 x) - cos^{2} (2 x))

sin^{2} (2 x) - cos^{4} (2 x)

Apply exponent rule:

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

sin (2 x) + cos^{2} (2 x) = 0, 0 \leq x \leq π : x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

sin (2 x) + cos^{2} (2 x) = 0, 0 \leq x \leq π

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

1 + sin (2 x) - sin^{2} (2 x) = 0

Solve by substitution

1 + sin (2 x) - sin^{2} (2 x) = 0

Let:

sin (2 x) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

Write in the standard form

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

- u^{2} + u + 1 = 0

Solve with the quadratic formula

- u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Substitute back

u = sin (2 x)

sin (2 x) = - \frac{- 1 + 5}{2}, sin (2 x) = \frac{1 + 5}{2}

sin (2 x) = - \frac{- 1 + 5}{2}, sin (2 x) = \frac{1 + 5}{2}

sin (2 x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq π : x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

sin (2 x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq π

Apply trig inverse properties

sin (2 x) = - \frac{- 1 + 5}{2}

General solutions for

sin (2 x) = - \frac{- 1 + 5}{2}

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

2 x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, 2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

2 x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, 2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

Solve

2 x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn : x = - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

2 x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn

Simplify

arcsin (- \frac{- 1 + 5}{2}) + 2 πn : - arcsin (\frac{5 - 1}{2}) + 2 πn

arcsin (- \frac{- 1 + 5}{2}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{5 - 1}{2}) = - arcsin (\frac{5 - 1}{2})

= - arcsin (\frac{5 - 1}{2}) + 2 πn

2 x = - arcsin (\frac{5 - 1}{2}) + 2 πn

Divide both sides by

2

2 x = - arcsin (\frac{5 - 1}{2}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + \frac{2 πn}{2}

Simplify

x = - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

x = - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

Solve

2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

Divide both sides by

2

2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

Solutions for the range

0 \leq x \leq π

x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

sin (2 x) = \frac{1 + 5}{2}, 0 \leq x \leq π :

No Solution

sin (2 x) = \frac{1 + 5}{2}, 0 \leq x \leq π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

sin (2 x) - cos^{2} (2 x) = 0, 0 \leq x \leq π : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

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

Rewrite using trig identities

- cos^{2} (2 x) + sin (2 x)

Use the Pythagorean identity:

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

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

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

- (1 - sin^{2} (2 x)) : - 1 + sin^{2} (2 x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

- 1 + sin (2 x) + sin^{2} (2 x) = 0

Solve by substitution

- 1 + sin (2 x) + sin^{2} (2 x) = 0

Let:

sin (2 x) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

Write in the standard form

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = sin (2 x)

sin (2 x) = \frac{- 1 + 5}{2}, sin (2 x) = \frac{- 1 - 5}{2}

sin (2 x) = \frac{- 1 + 5}{2}, sin (2 x) = \frac{- 1 - 5}{2}

sin (2 x) = \frac{- 1 + 5}{2}, 0 \leq x \leq π : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

sin (2 x) = \frac{- 1 + 5}{2}, 0 \leq x \leq π

Apply trig inverse properties

sin (2 x) = \frac{- 1 + 5}{2}

General solutions for

sin (2 x) = \frac{- 1 + 5}{2}

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

2 x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, 2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

2 x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, 2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Solve

2 x = arcsin (\frac{- 1 + 5}{2}) + 2 πn : x = \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

2 x = arcsin (\frac{- 1 + 5}{2}) + 2 πn

Divide both sides by

2

2 x = arcsin (\frac{- 1 + 5}{2}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

Solve

2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Divide both sides by

2

2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

Solutions for the range

0 \leq x \leq π

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

sin (2 x) = \frac{- 1 - 5}{2}, 0 \leq x \leq π :

No Solution

sin (2 x) = \frac{- 1 - 5}{2}, 0 \leq x \leq π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

Combine all the solutions

x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}, x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (2 x) = cos (2 x)

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

Check the solution

\frac{π + arcsin ( \frac{5 - 1}{2} )}{2} :

False

\frac{π + arcsin ( \frac{5 - 1}{2} )}{2}

Plug in

n = 1

\frac{π + arcsin ( \frac{5 - 1}{2} )}{2}

For

tan (2 x) = cos (2 x)

plug in

x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}

tan 2 \cdot \frac{π + arcsin ( \frac{5 - 1}{2} )}{2} = cos 2 \cdot \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}

Refine

0.78615 \dots = - 0.78615 \dots

\Rightarrow False

Check the solution

\frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2} :

False

\frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

Plug in

n = 1

\frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

For

tan (2 x) = cos (2 x)

plug in

x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

tan 2 \cdot \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2} = cos 2 \cdot \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

Refine

- 0.78615 \dots = 0.78615 \dots

\Rightarrow False

Check the solution

\frac{arcsin ( \frac{5 - 1}{2} )}{2} :

True

\frac{arcsin ( \frac{5 - 1}{2} )}{2}

Plug in

n = 1

\frac{arcsin ( \frac{5 - 1}{2} )}{2}

For

tan (2 x) = cos (2 x)

plug in

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}

tan 2 \cdot \frac{arcsin ( \frac{5 - 1}{2} )}{2} = cos 2 \cdot \frac{arcsin ( \frac{5 - 1}{2} )}{2}

Refine

0.78615 \dots = 0.78615 \dots

\Rightarrow True

Check the solution

\frac{π - arcsin ( \frac{5 - 1}{2} )}{2} :

True

\frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

Plug in

n = 1

\frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

For

tan (2 x) = cos (2 x)

plug in

x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

tan 2 \cdot \frac{π - arcsin ( \frac{5 - 1}{2} )}{2} = cos 2 \cdot \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

Refine

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow True

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

Show solutions in decimal form

x = \frac{0.66623 \dots}{2}, x = \frac{π - 0.66623 \dots}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)=cos(2x),0<= x<= pi ?
The general solution for tan(2x)=cos(2x),0<= x<= pi is x=(0.66623…)/2 ,x=(pi-0.66623…)/2