What is the general solution for cos^2(x)-sin^2(x)=2-5cos(x) ?

The general solution for cos^2(x)-sin^2(x)=2-5cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(x)-sin^2(x)=2-5cos(x)

Solution

cos^{2} (x) - sin^{2} (x) = 2 - 5 cos (x)

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - sin^{2} (x) = 2 - 5 cos (x)

Subtract

2 - 5 cos (x)

from both sides

cos^{2} (x) - sin^{2} (x) - 2 + 5 cos (x) = 0

Rewrite using trig identities

- 2 + cos^{2} (x) - sin^{2} (x) + 5 cos (x)

Use the Pythagorean identity:

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

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

= - 2 + cos^{2} (x) - (1 - cos^{2} (x)) + 5 cos (x)

Simplify

- 2 + cos^{2} (x) - (1 - cos^{2} (x)) + 5 cos (x) : 2 cos^{2} (x) + 5 cos (x) - 3

- 2 + cos^{2} (x) - (1 - cos^{2} (x)) + 5 cos (x)

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

= - 2 + cos^{2} (x) - 1 + cos^{2} (x) + 5 cos (x)

Simplify

- 2 + cos^{2} (x) - 1 + cos^{2} (x) + 5 cos (x) : 2 cos^{2} (x) + 5 cos (x) - 3

- 2 + cos^{2} (x) - 1 + cos^{2} (x) + 5 cos (x)

Group like terms

= cos^{2} (x) + cos^{2} (x) + 5 cos (x) - 2 - 1

Add similar elements:

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

= 2 cos^{2} (x) + 5 cos (x) - 2 - 1

Subtract the numbers:

- 2 - 1 = - 3

= 2 cos^{2} (x) + 5 cos (x) - 3

= 2 cos^{2} (x) + 5 cos (x) - 3

= 2 cos^{2} (x) + 5 cos (x) - 3

- 3 + 2 cos^{2} (x) + 5 cos (x) = 0

Solve by substitution

- 3 + 2 cos^{2} (x) + 5 cos (x) = 0

Let:

cos (x) = u

- 3 + 2 u^{2} + 5 u = 0

- 3 + 2 u^{2} + 5 u = 0 : u = \frac{1}{2}, u = - 3

- 3 + 2 u^{2} + 5 u = 0

Write in the standard form

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

2 u^{2} + 5 u - 3 = 0

Solve with the quadratic formula

2 u^{2} + 5 u - 3 = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = - 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 3) = 7

5^{2} - 4 \cdot 2 (- 3)

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 \cdot 2}, u_{2} = \frac{- 5 - 7}{2 \cdot 2}

u = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

\frac{- 5 + 7}{2 \cdot 2}

Add/Subtract the numbers:

- 5 + 7 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 5 - 7}{2 \cdot 2} : - 3

\frac{- 5 - 7}{2 \cdot 2}

Subtract the numbers:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 12}{4}

Apply the fraction rule:

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

= - \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= - 3

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 3

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

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

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}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 3 :

No Solution

cos (x) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Graph

Popular Examples

tan(x/4)=-sqrt(3)

tan (\frac{x}{4}) = - 3

8sin^2(x)-2sin(x)-1=0,-pi/2 <x< pi/2

8 sin^{2} (x) - 2 sin (x) - 1 = 0, - \frac{π}{2} < x < \frac{π}{2}

cos(4x)-cos(3x)=0

cos (4 x) - cos (3 x) = 0

2cos^2(A)+cos(A)-1=0

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

tan^3(x)=tan(x),0<= x<= 2pi

tan^{3} (x) = tan (x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-sin^2(x)=2-5cos(x) ?
The general solution for cos^2(x)-sin^2(x)=2-5cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin