What is the general solution for 2cos(x)-tan(x)=0 ?

The general solution for 2cos(x)-tan(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin

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Popular Trigonometry >

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

Solution

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

Solution

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Degrees

x = 51.33171 \dots^{\circ} + 36 0^{\circ} n, x = 128.66828 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Express with sin, cos

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

2 cos (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

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

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

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

Add

sin (x)

to both sides

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

Square both sides

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

Subtract

sin^{2} (x)

from both sides

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

Factor

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

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

Rewrite

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

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

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

Rewrite

4

2^{2}

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

2 cos^{2} (x) + sin (x) = 0 : x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{- 1 + 17}{4}, u = \frac{1 + 17}{4}

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

Expand

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

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

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

Expand

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

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 2

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

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

1^{2} - 4 (- 2) \cdot 2 = 17

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 2

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 17}{2 ( - 2 )}, u_{2} = \frac{- 1 - 17}{2 ( - 2 )}

u = \frac{- 1 + 17}{2 ( - 2 )} : - \frac{- 1 + 17}{4}

\frac{- 1 + 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 17}{- 4}

Apply the fraction rule:

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

= - \frac{- 1 + 17}{4}

u = \frac{- 1 - 17}{2 ( - 2 )} : \frac{1 + 17}{4}

\frac{- 1 - 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 17}{- 4}

Apply the fraction rule:

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

- 1 - 17 = - (1 + 17)

= \frac{1 + 17}{4}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 17}{4}, u = \frac{1 + 17}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + 17}{4}, sin (x) = \frac{1 + 17}{4}

sin (x) = - \frac{- 1 + 17}{4}, sin (x) = \frac{1 + 17}{4}

sin (x) = - \frac{- 1 + 17}{4} : x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = - \frac{- 1 + 17}{4}

Apply trig inverse properties

sin (x) = - \frac{- 1 + 17}{4}

General solutions for

sin (x) = - \frac{- 1 + 17}{4}

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

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{1 + 17}{4} :

No Solution

sin (x) = \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

2 cos^{2} (x) - sin (x) = 0 : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

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

Expand

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

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

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

Expand

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

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 2

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

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

Apply rule

- (- a) = a

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

(- 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 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

Apply the fraction rule:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

Apply the fraction rule:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4} :

No Solution

sin (x) = - \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{17 - 1}{4} : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

sin (x) = \frac{17 - 1}{4}

Apply trig inverse properties

sin (x) = \frac{17 - 1}{4}

General solutions for

sin (x) = \frac{17 - 1}{4}

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

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

arcsin (- \frac{- 1 + 17}{4}) + 2 πn :

False

arcsin (- \frac{- 1 + 17}{4}) + 2 πn

Plug in

n = 1

arcsin (- \frac{- 1 + 17}{4}) + 2 π 1

For

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

plug in

x = arcsin (- \frac{- 1 + 17}{4}) + 2 π 1

2 cos (arcsin (- \frac{- 1 + 17}{4}) + 2 π 1) - tan (arcsin (- \frac{- 1 + 17}{4}) + 2 π 1) = 0

Refine

2.49924 \dots = 0

\Rightarrow False

Check the solution

π + arcsin (\frac{- 1 + 17}{4}) + 2 πn :

False

π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

Plug in

n = 1

π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1

For

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

plug in

x = π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1

2 cos (π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1) - tan (π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1) = 0

Refine

- 2.49924 \dots = 0

\Rightarrow False

Check the solution

arcsin (\frac{17 - 1}{4}) + 2 πn :

True

arcsin (\frac{17 - 1}{4}) + 2 πn

Plug in

n = 1

arcsin (\frac{17 - 1}{4}) + 2 π 1

For

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

plug in

x = arcsin (\frac{17 - 1}{4}) + 2 π 1

2 cos (arcsin (\frac{17 - 1}{4}) + 2 π 1) - tan (arcsin (\frac{17 - 1}{4}) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

Check the solution

π - arcsin (\frac{17 - 1}{4}) + 2 πn :

True

π - arcsin (\frac{17 - 1}{4}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{17 - 1}{4}) + 2 π 1

For

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

plug in

x = π - arcsin (\frac{17 - 1}{4}) + 2 π 1

2 cos (π - arcsin (\frac{17 - 1}{4}) + 2 π 1) - tan (π - arcsin (\frac{17 - 1}{4}) + 2 π 1) = 0

Refine

0 = 0

\Rightarrow True

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)-tan(x)=0 ?
The general solution for 2cos(x)-tan(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin