What is the general solution for cos^2(2x)-3cos(2x)=0 ?

The general solution for cos^2(2x)-3cos(2x)=0 is x= pi/4+pin,x=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(2x)-3cos(2x)=0

Solution

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

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (2 x) = u

u^{2} - 3 u = 0

u^{2} - 3 u = 0 : u = 3, u = 0

u^{2} - 3 u = 0

Solve with the quadratic formula

u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

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

n

is even

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

= 3^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} - 0

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

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 3 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 3 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 3, u = 0

Substitute back

u = cos (2 x)

cos (2 x) = 3, cos (2 x) = 0

cos (2 x) = 3, cos (2 x) = 0

cos (2 x) = 3 :

No Solution

cos (2 x) = 3

- 1 \leq cos (x) \leq 1

No Solution

cos (2 x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos (2 x) = 0

General solutions for

cos (2 x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = \frac{π}{2} + 2 πn, 2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{π}{2} + 2 πn, 2 x = \frac{3 π}{2} + 2 πn

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Popular Examples

csc^2(x)=sin(x)

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

cos(θ)= 4/7 ,tan(θ)<0

cos (θ) = \frac{4}{7}, tan (θ) < 0

tan(θ)= 4/2

tan (θ) = \frac{4}{2}

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

e^{s i n (x)} - e^{- s i n (x)} = 2

cos(x)*sec(x)*cot^2(x)=csc^2(x)

cos (x) \cdot sec (x) \cdot cot^{2} (x) = csc^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(2x)-3cos(2x)=0 ?
The general solution for cos^2(2x)-3cos(2x)=0 is x= pi/4+pin,x=(3pi)/4+pin