What is the general solution for 1+2cos(x)+cos^2(x)=3sin^2(x) ?

The general solution for 1+2cos(x)+cos^2(x)=3sin^2(x) is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin

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1+2cos(x)+cos^2(x)=3sin^2(x)

Solution

1 + 2 cos (x) + cos^{2} (x) = 3 sin^{2} (x)

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + 2 cos (x) + cos^{2} (x) = 3 sin^{2} (x)

Subtract

3 sin^{2} (x)

from both sides

1 + 2 cos (x) + cos^{2} (x) - 3 sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

1 + cos^{2} (x) + 2 cos (x) - 3 (1 - cos^{2} (x)) : 4 cos^{2} (x) + 2 cos (x) - 2

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

Expand

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

- 3 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos^{2} (x)

= - 3 \cdot 1 - (- 3) cos^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 cos^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

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

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

Simplify

1 + cos^{2} (x) + 2 cos (x) - 3 + 3 cos^{2} (x) : 4 cos^{2} (x) + 2 cos (x) - 2

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

Group like terms

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

Add similar elements:

cos^{2} (x) + 3 cos^{2} (x) = 4 cos^{2} (x)

= 4 cos^{2} (x) + 2 cos (x) + 1 - 3

Add/Subtract the numbers:

1 - 3 = - 2

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 2, c = - 2

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

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

2^{2} - 4 \cdot 4 (- 2) = 6

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 2 + 6}{2 \cdot 4}, u_{2} = \frac{- 2 - 6}{2 \cdot 4}

u = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

\frac{- 2 + 6}{2 \cdot 4}

Add/Subtract the numbers:

- 2 + 6 = 4

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 2 - 6}{2 \cdot 4} : - 1

\frac{- 2 - 6}{2 \cdot 4}

Subtract the numbers:

- 2 - 6 = - 8

= \frac{- 8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 1+2cos(x)+cos^2(x)=3sin^2(x) ?
The general solution for 1+2cos(x)+cos^2(x)=3sin^2(x) is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin