What is the general solution for cos(2x)=tan(2x) ?

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

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cos(2x)=tan(2x)

Solution

cos (2 x) = tan (2 x)

Solution

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

Degrees

x = 19.08635 \dots^{\circ} + 18 0^{\circ} n, x = 70.91364 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x) = tan (2 x)

Subtract

tan (2 x)

from both sides

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

Express with sin, cos

cos (2 x) - tan (2 x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

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

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

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

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

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

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{5 - 1}{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

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 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, - (- 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{5 - 1}{2}

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

Remove parentheses:

(- a) = - a, - (- 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 = - (5 - 1)

= \frac{5 - 1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (2 x)

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

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Solve

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

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

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

Combine all the solutions

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

Show solutions in decimal form

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

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Frequently Asked Questions (FAQ)

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