English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4cos(2x)=4cos^2(x)-1

Solution

4 cos (2 x) = 4 cos^{2} (x) - 1

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (2 x) = 4 cos^{2} (x) - 1

Subtract

4 cos^{2} (x) - 1

from both sides

4 cos (2 x) - 4 cos^{2} (x) + 1 = 0

Rewrite using trig identities

1 + 4 cos (2 x) - 4 cos^{2} (x)

Use the Double Angle identity:

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

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

= 1 + 4 cos (2 x) - 4 (cos (2 x) + sin^{2} (x))

Simplify

1 + 4 cos (2 x) - 4 (cos (2 x) + sin^{2} (x)) : 1 - 4 sin^{2} (x)

1 + 4 cos (2 x) - 4 (cos (2 x) + sin^{2} (x))

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = - 4, b = cos (2 x), c = sin^{2} (x)

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

Apply minus-plus rules

+ (- a) = - a

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

= 1 + 4 cos (2 x) - 4 cos (2 x) - 4 sin^{2} (x)

Add similar elements:

4 cos (2 x) - 4 cos (2 x) = 0

= 1 - 4 sin^{2} (x)

= 1 - 4 sin^{2} (x)

1 - 4 sin^{2} (x) = 0

Solve by substitution

1 - 4 sin^{2} (x) = 0

Let:

sin (x) = u

1 - 4 u^{2} = 0

1 - 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

1 - 4 u^{2} = 0

Move

1

to the right side

1 - 4 u^{2} = 0

Subtract

1

from both sides

1 - 4 u^{2} - 1 = 0 - 1

Simplify

- 4 u^{2} = - 1

- 4 u^{2} = - 1

Divide both sides by

- 4

- 4 u^{2} = - 1

Divide both sides by

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 1}{- 4}

Simplify

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Popular Examples

tan(8b)=cot(10b)

tan (8 b) = cot (10 b)

solvefor θ,cos(θ)=0 solve for

θ, cos (θ) = 0

sec^2(2x)-2tan(2x)=0

sec^{2} (2 x) - 2 tan (2 x) = 0

solvefor x,du=cos(4x)solve for

x, d u = cos (4 x)

((sin(θ))/(2cos(θ)))-1=0

(\frac{sin ( θ )}{2 cos ( θ )}) - 1 = 0